Calculus - II Multivariable Calculus. M.Thamban Nair. Department of Mathematics Indian Institute of Technology Madras

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1 Calculus - II Multivariable Calculus M.Thamban Nair epartment of Mathematics Indian Institute of Technology Madras February 27 Revised: January 29

2 Contents Preface v 1 Functions of everal Variables Introduction Geometric Representation of Functions equences and their Convergence Limit and Continuity ome More Topological Notions Two Theorems Partial erivatives Partial Increments and Total Increment Total ifferential, Gradient, ifferentiability erivatives of Composition of Functions erivatives of Implicitly efined Functions Level Curve and Level urface irectional erivatives Taylor s Formula Maxima and Minima Method of Lagrange Multipliers Problems Multiple Integrals ouble Integrals Calculating double integrals ii

3 Contents iii Calculating double integrals using polar coordinates Multiple integrals using change of variables urfaces and their representations Area of a surface using double integrals Mass, Moment of inertia and Centre of gravity (i) Mass: (ii) Moment of inertia: (iii) Centre of gravity: Triple Integrals Calculating triple integrals Applications Line Integrals Curves in the Plane and pace Line Integrals Motivation and definition Evaluation Area of a domain using line integral Line integral of scalar valued functions Green s Theorem Conditions for line integral to be independent of path of integration urface Integral Integral over urfaces Examples tokes Theorem Examples Gauss s ivergence Theorem Examples

4 iv Contents 5 Appendix 86 Index 87

5 Preface These notes are primarily for the personal use of teachers and students of IIT Madras. M.Thamban Nair v

6 1 Functions of everal Variables 1.1 Introduction Functions of more than one variables come naturally in applications. For example, in physics one comes across the relation P V T = c, constant, where P, V, T represents the pressure, volume and temperature of an ideal gas. ince P = ct V, V = ct P, T = P V c each of P, V, T can be thought of as a function of the remaining two variables. In elementary geometry, we know that the area of a triangle of base length b and altitude h, area of a rectangle of sides a and b, and area of a circle of radius r are given by 1 2 bh, ab and πr2, respectively, and they are functions of the variables (b, h), (a, b) and r, respectively. Also, distance of a point (x, y) in the plane from the origin (, ) is given by x 2 + y 2. We shall introduce some notations and basic definitions: We use the symbol N to denote the set of all positive integers and R to denote the set of all real numbers. For k N, we denote by R k the set of all k-tuples (x 1,..., x k ) with x i R for i {1,..., k}. Also, for u = (x 1,..., x k ) R k, we denote by u the positive square root of x x , x2 k, i.e., u := x x , x2 k. Note that, if u and v are points in R 2 or in R 3, then u v is the distance between u and v. efinition 1.1 By a function of several variables we mean a function f : R, where is a subset of R k for some k {2, 3,...}. We write this fact by z = f(x 1,..., x k ), (x 1,..., x k ), 1

7 2 Functions of everal Variables and say that z is a function of (x 1,..., x k ). The set is called the domain of the function f, and we say that f is defined on. In this course, for the sake of simplicity of presentation, we shall consider functions of two variables, i.e., R 2. Most of the results that we study for two variables can be extended to the case of more than two variables. 1.2 Geometric Representation of Functions Geometrically a function of two variables represents a surface in the 3-dimensional space, in such a way that the projection of to the xy-plane is the domain, and each line parallel to the z-axis and passing through a point in cuts the surface at one and only one point. In fact, the surface corresponding to the function f is the graph of f. Thus, given a function f : R, the corresponding surface is the set of all points (x, y, z) R 3 such that z = f(x, y) with (x, y), i.e., = {(x, y, z) R 3 : z = f(x, y), (x, y) }. EXAMPLE 1.1 Here are two examples functions of two variables: (a) z = f(x, y) := 1 x 2 y 2, := {(x, y) : x 2 + y 2 1}. (b) z = f(x, y) := xy x 2 + y 2, := {(x, y) : x2 + y 2 }. efinition 1.2 A subset of R 2 is said to be a bounded set if there exits M > such that u M for all u. ets which are not bounded are called unbounded. In Example 1.1, the domain in (a) is bounded and the domain in (b) is unbounded. 1.3 equences and their Convergence efinition 1.3 A sequence in R 2 is a function from N to R 2. If f is a sequence in R 2, and if f(n) = (x n, y n ), then the sequence f is also written as {(x n, y n )} or {u n } where u n := (x n, y n ) for n N. efinition 1.4 A sequence (u n ) in R 2 is said to converge to a point u R 2 if u n u as n, i.e., ( x n x 2 + y n y 2 ) 1/2 as n, where u n = (x n, y n ) and u = (x, y), and in that case we say that u is the limit of (u n ), and write u n u or lim n u n = u.

8 Limit and Continuity 3 A sequence (u n ) in R 2 is said to be a convergent sequence if it converges to some point in R 2. Thus, a sequence (u n ) in R 2 converges to u R 2 if and only if for every ε >, there exists N N such that u n u < ε whenever n N. Remark 1.1 ifferent sequences may have the same limit. For example, the sequences (u n ) and (v n ) with u n := ( n n+1, ) and v n := ( n n+1, n) 1 have the same limit u := (1, ). Exercise 1.1 how that the limit of a convergent sequence is unique. Exercise 1.2 Let (u n ) be a sequence in R 2 with u n := (x n, y n ) for n N and u := (x, y) R 2. how that (i) u n u if and only if x n x and y n y, and (ii) u n u if and only if for every ε >, there exists N N such that x n x < ε and y n y < ε whenever n N. Exercise 1.3 Let (u n ) be a sequence in R 2. how that (u n ) converges if and only if for every ε >, there exists a positive integer N such that u n u m < ε for every n, m N. 1.4 Limit and Continuity We shall discuss limit, continuity and differentiability of functions of several variables. One of the primary concepts required to do these is that of a neighbourhood of a point in R 2. efinition 1.5 By a neighbourhood of a point u = (x, y ) R 2 we mean the set of all points u = (x, y) R 2 such that u u < δ for some δ >. uch a set, {u R 2 : u u < δ} is called a δ-neighbourhood of u = (x, y ), or an open disc with centre u and radius δ. A δ-neighbourhood of a point u is usually denoted by B δ (u ). efinition 1.6 A neighbourhood with its centre u deleted is called a deleted neighbourhood of u. Thus, {u R 2 : < u u < δ} is a deleted neighbourhood of u. efinition 1.7 A point u R 2 is called a limit point of R 2, if every deleted neighbourhood of u contains atleast one point from. Exercise 1.4 how that a point u R 2 is a limit point of R 2 if and only if there exists a sequence (u n ) of distinct points in which converges to u.

9 4 Functions of everal Variables EXAMPLE 1.2 Let := {(x, y) R 2 : x 2 + y 2 < 1}. Then u := (1, ) is a limit point of, as the sequence (u n ) with u n := ( n n+1, ) converges to u. Also, the sequence (v n ) with v n := ( n n+1, n) 1 converges to u. In fact every point in the closed disc {(x, y) R 2 : x 2 + y 2 1} is a limit point of. efinition 1.8 uppose f is defined on a set R 2, and u = (x, y ) R 2 be a limit point of. We say that f has the limit α as u = (x, y) approaches u if for every ε > there exists a δ > such that We write the above fact by f(u) α < ε whenever u, < u u < δ. lim f(u) = α or lim f(x, y) = α. u u (x,y) (x,y ) Thus, for function f defined on a set, if u is a limit point of, then lim u u f(u) = α if and only if for every ε >, there exists a deleted neighbourhood of u such that u = f(u) (α ε, α + ε). Proposition 1.1 Let f : R, and for some limit point u of, lim f(u) exist. u u If α = lim f(u), then for every sequence (u n ) in with u n u as n, we u u have f(u n ) α as n. Proof. Let α = lim u u f(u). Then, for every ε >, there exists δ > such that for every u with u u < δ we have f(u) α < ε. Let (u n ) be a sequence in such that u n u as n. Then, there exists n N such that u n u < δ for all n n. Therefore, we have f(u n ) α < ε for all n n. Thus, f(u n ) α as n. Remark 1.2 By Proposition 1.1, if (u n ) and v n ) are sequences in having the same limit u, but their images under f have different limits, then lim f(u) does u u not exist. EXAMPLE 1.3 Let f(x, y) = 1 x 2 y 2, := {(x, y) : x 2 + y 2 1}. Then f(x, y) = 1. Let us show this: Let ε > be given. If ε 1, then f(u) 1 = lim (x,y) (,) 1 f(u) < ε for all u. o, let < ε < 1. Then we have f(u) 1 < ε 1 ε < f(u) (1 ε) 2 < 1 u 2 u 2 < 1 (1 ε) 2. Thus, taking δ := 1 (1 ε) 2 = 2ε ε 2 >, we have f(u) 1 < ε whenever u < δ.

10 Limit and Continuity 5 Exercise 1.5 how that the function f in Example 1.3 is continuous at all points in := {(x, y) : x 2 + y 2 1}. EXAMPLE 1.4 Let f(x, y) := xy x2 y 2 x 2 + y 2, := {(x, y) : x2 + y 2 }. taking x = r cos θ and y = r sin θ we have r 2 = x 2 + y 2, and f(x, y) = r 2 sin 4θ 4 r2 4 as r2. Then In fact, for ε >, Thus, f(x, y) u 2 4 < ε whenever u < 2 ε. lim f(x, y) =. Note that f is not defined at (, ). (x,y) (,) If we define f(x, y) := { f(x, y), (x, y) (, ),, (x, y) = (, ), then f is continuous at every point in R 2. EXAMPLE 1.5 Let f(x, y) := xy x 2 + y 2, := {(x, y) : x2 + y 2 }. Then f(x, y) does not exist. To see this, for each m R, consider the straight lim (x,y) (,) line L m := {(x, y) : y = mx}, i.e., the straight line passing through the origin with slope m. Then we see that for (x, y) L m, f(x, y) = m 1 + m 2. m Let α be any real number and m is such that α. If we take ε > is 1 + m2 such that ε < α m 1 + m 2, then we see that there does not exist a δ > satisfying f(u) α < ε for all u with u < δ. Thus, α is not a limit of f at. This is true for any α. Hence, f does not have a limit at. In fact, we can find different sequences (u n ) and (v n ) in having the same limit (, ), but (f(u n )) and (f(v n ) have different limits. x2 y EXAMPLE 1.6 Let z = f(x, y) := x 4 + y 2, := {(x, y) : x2 + y 2 }. Then lim f(x, y) does not exist. To see this, for each m R, consider the set (x,y) (,) A m := {(x, y) : y = mx 2 }. Then we see that for (x, y) A m, f(x, y) = m 1 + m 2. Again, by the same argument as in last example, the function does not have limit at (, ). Remark 1.3 It is to be observed that the limit defined above is quite different from the limits: lim lim f(x, y), lim lim f(x, y). y y x x x x y y

11 6 Functions of everal Variables It is possible that the limit lim f(x, y) does not exist but one or both of the limits (x,y) (x,y ) lim lim f(x, y), lim lim f(x, y) exist, and y y x x x x y y any one or both of lim lim f(x, y) exists. (x,y) (x,y ) x x lim f(x, y) and lim y y y y lim f(x, y) may not exist, but x x To illustrate the statements in the above remark we consider a two examples. EXAMPLE 1.7 Let f(x, y) := Then we see that and lim lim y f(x, y) = lim y x y y = 1, (y x)(1 + x), := {(x, y) : x + y, y 1}. (y + x)(1 + y) lim lim f(x, y) = lim ( 1)(1 + x) = 1, x y x m 1 lim f(x, y) = y=mx,x m + 1. Thus, separate limit exit, but the limit does not exists at (, ). Note that the above function is not defined in a deleted neighbourhood of (, ). EXAMPLE 1.8 Let f(x, y) := x sin 1 y + y sin 1, := {(x, y) : xy }. x Then we see that separate limit do not exist at (, ), but f(x, y) x + y so that lim f(x, y) =. (x,y) (,) efinition 1.9 A function f defined on a set R 2 is said to be continuous at a point u if lim f(u) exists and is equal to f(u ). u u Remark 1.4 We note that in order to define limit of a function f at a point u = (x, y ), it is not necessary that the function is defined at u, whereas to define continuity of f at u it is necessary that u belongs to the domain of f. Remark 1.5 By Proposition 1.1, if f : R is continuous at u, then for very sequence (u n ) which converges to u, we have f(u n ) f(u ). Thus, if there exists a sequence (u n ) with limit u such that f(u n ) f(u ), then f is not continuous at u.

12 Limit and Continuity 7 In fact, the converse of this statement is also true. That is, for every convergent sequence (u n ) with limit u if we have lim f(u n) = f(u ), then f is continuous at n u. Exercise 1.6 Prove the last statement in he above remark. Remark 1.6 (i) uppose a function f is not defined at a point u, but lim u u f(u) exists. Then we may extend the function f to a new function f defined on := {u }, where is the domain of f, by { f(u), u, f(u) := lim f(u), u = u. u u Then we have lim u u f(u) = f(u ). Thus, f is continuous at u. (ii) uppose a function f does not have a limit at a point (x, y ), and suppose that (x, y ) is not in the domain of definition of f. Then, no matter whatever value we assign at (x, y ), the extended function cannot be continuous. EXAMPLE 1.9 In Example 1.3, the function f is continuous at every point in its domain of definition. EXAMPLE 1.1 In Example 1.4, the function is not defined at the point u = (, ). But, the extended function is continuous at every point in R 2. f(x, y) := { f(x, y), (x, y) (, ),, (x, y) = (, ), EXAMPLE 1.11 In Examples 1.5 and 1.6, the functions cannot be redefined at u = (, ) so as to make them continuous at u ome More Topological Notions For stating two important theorems concerning continuous functions, we need to use a few more definitions. Although we state them for subsets of R 2, they are equally valid for subsets of R k as well for any k N. efinition 1.1 A point u = (x, y ) R 2 is said to be an interior point of a set R 2 if contains a neighbourhood of u, that is, if there exists δ > such that B δ (u ). efinition 1.11 A subset G of R 2 is said to be an open set if every point in G is an interior point of G.

13 8 Functions of everal Variables efinition 1.12 A point u = (x, y ) is said to be a boundary point of a set R 2 if every neighbourhood of u contains points from and c. The set of all boundary points of a set is called the boundary of. efinition 1.13 A subset F of R 2 is said to be a closed set if F contains all its limit points. Exercise 1.7 Let R 2. Prove that the following are equivalent: (i) is closed. (ii) contains all its boundary points. (iii) c is open in R 2. Exercise 1.8 Let R 2. Prove the: (i) Every δ- neighbourhood of a point is an open set. (ii) A point u is an interior point of iff there exists an open set G such that u G. (iii) A point u R 2 is a limit point of iff for open set G with u G, G ( \ {u }). (iv) A point u R 2 is a boundary point of iff for open set G with u G, G and G c. EXAMPLE 1.12 The following statements can be easily verified. (i) The set 1 := {(x, y) R 2 : x 2 + y 2 < 1} is an open set, but not a closed set. (ii) The set 2 := {(x, y) R 2 : x 2 + y 2 1} is a closed set, but not an open set. (iii) The set 3 := {(x, y) R 2 : x 2 + y 2 < 1} {(1, )} is neither open nor closed in R 2. (iv) The set 1 := {(x, y) R 2 : x 2 + y 2 = 1} is the boundary of the sets 1, 2 and 3. (v) If 4 := {(x, y) R 2 : < x 2 + y 2 < 1}, then 4 is neither open nor closed, and 2 is the boundary of 4. (v) If 5 is the set of all points in 1 with rational coordinates, then 5 is neither open nor closed, and 2 is the boundary of 5. efinition 1.14 Let ϕ : [a, b] be a function with ϕ(t) = (f(t), g(t)), t [a, b], where f and g are real valued functions defined on [a, b]. Then ϕ : [a, b] is said to be continuous at t [a, b] if both f and g are continuous at t. efinition 1.15 Let R 2. By a path in we mean a continuous function γ : [, 1] R 2 with γ(t) for all t [, 1]. The point u := γ() is called the

14 Limit and Continuity 9 initial point of the path γ and u 1 := γ(1) is called the final or terminal point of γ, and we say that γ is a path joining u to u 1. efinition 1.16 A subset of R 2 is said to be a connected set if any two points in can be joined by a path in, that is, for any u and u 1 in, there exists a curve γ : [, 1] in such that u = γ() and u 1 := γ(1). We observe that if any two points in R 2 can be joined by a polygonal line, then is connected. EXAMPLE 1.13 The sets 1, 2, 3, 4, 1 in Example 1.12 are connected, whereas the set 5 is not connected. efinition 1.17 An open connected set together with, possibly, some or all of its boundary points is called a domain. Remark 1.7 Conventionally, a domain is an open connected set. We digressed from this convention to be in conformity with most of the engineering mathematics books. Thus, R 2 is a domain if and only = G E where E is either empty set or a set of some of the boundary points of G Two Theorems Theorem 1.2 uppose f is a continuous function defined on a closed and bounded domain R 2. Then we have following: (a) (On attaining maximum and minimum) There exist points (x 1, y 1 ) and (x 2, y 2 ) in such that f(x 1, y 1 ) f(x, y) f(x 2, y 2 ) (x, y), and in that case we write f(x 1, y 1 ) = min f(x, y), x f(x 2, y 2 ) = max f(x, y). x (b) (Intermediate value theorem) If (x 1, y 1 ) and (x 2, y 2 ) are in and c R is such that f(x 1, y 1 ) < c < f(x 2, y 2 ) then there exists (x, y ) such that f(x, y ) = c.

15 1 Functions of everal Variables 1.5 Partial erivatives efinition 1.18 uppose f is a (real valued) function defined in a neigbourhood of a point (x, y ). Then f is said to have the partial derivative with respect d to x at (x, y ) if dx f(x, y ) exists at x, and it is denoted by f x (x, y ). Thus, if f has partial derivative with respect to x at (x, y ), then f x (x, y ) = d dx f(x, y ) x=x = lim x f(x + x, y ) f(x, y ), x and it is called the partial derivative of f with respect to x at (x, y ). imilarly, d if if dy f(x, y) exists at y, then the partial derivative of f with respect to y at (x, y ), denoted by f y (x, y ), is defined by f y (x, y ) = d dy f(x, y) y=y = lim y f(x, y + y) f(x, y ). y Partial derivatives f x (x, y ) and f y (x, y ) are also denoted by f x (x, y ) and f y (x, y ) respectively. We denote f xx (x, y ) := (f x ) x (x, y ), f xy (x, y ) := (f x ) y (x, y ), Thus, f yx (x, y ) := (f y ) x (x, y ), f yy (x, y ) := (f y ) y (x, y ). f xx (x, y ) := d dx f x(x, y ) x=x, f yx (x, y ) := d dx f x(x, y ) x=x, f xy (x, y ) := d dy f x(x, y) y=y, f yy (x, y ) := d dy f y(x, y) y=y. Remark 1.8 It should be kept in mind that f x (x, y ) cannot be interpreted as lim f x(x, y) even if the later limit exist. This can be explained in the case (x,y) (x,y ) of a function of single variable itself: Consider the function f(x) = { 1, x >, x. Clearly, f (x) = for every x so that lim x f (x) =, but, we can not say that f () = lim x f (x), as the function f is not even continuous at.

16 Partial erivatives 11 Also, there are situations in which f x (x, y ) exists but not exist. To see this consider the function { x f(x) = 2 sin(1/x), x, x =. lim f x(x, y) need (x,y) (x,y ) Then we have f(x) f() x = x 2 sin(1/x) as x so that f () =, but lim x f (x) = lim x [2x sin(1/x) cos(1/x)] does not exist. Partial derivatives of more than two variables are also defined analogously: For example if f is a function of three variables, say (x 1, x 2, x 3 ), then we can define f xi, f xi x j, f xi x j x k for i, j, k {1, 2, 3}. EXAMPLE 1.14 Let f(x, y) := x 2 y + y 3, (x, y) R 2. Then f x = 2xy; (f x ) y = 2x = (f y ) x ; f xx = 2y; f yy = 6y. EXAMPLE 1.15 Let u = e xy sin z. Then, u x = e xy y sin z; u y = e xy x sin z; u xy := (u x ) y = sin z(xy + 1)e xy ; u xyz := (u xy ) z = cos z(xy + 1)e xy ; u yz := (u y ) z = e xy x cos z; u yzx := (u yz ) x = cos z(xy + 1)e xy ; EXAMPLE 1.16 Let f(x, y) := xy x 2 for (x, y) (, ). We have already observed that this function does not have a limit at (, ). In order to check the + y2 existence of partial derivatives of a function at a point (x, y ), a necessary condition is that the function has to be defined at that point. o, first let us the extend the function f to { f(x, y), (x, y) (, ) g(x, y) := α, (x, y) = (, ) for some α R. Note that the functions g(x, ) and g(, y) are continuous at x = and y = respectively if and only if α =. o, let { f(x, y), (x, y) (, ) g(x, y) :=, (x, y) = (, ) Note that the g does not have a limit at (, ). However, g(x, ) = and g(, y) = for all x, y R. Hence, g x (x, ) =, g y (, y) = x, y R,

17 12 Functions of everal Variables g xx (x, ) =, g yy (, y) = x, y R. Let us see if g xy (, ) exists. Recall that g xy (, ) := d dy g x(, y) y=. We have to see whether the function g x (, y) is differentiable at y =. Note that for y, g( x, y) g(, y) g( x, y) y g x (, y) = lim = lim = lim x x x x x ( x) 2 + y 2 = 1 y. imilarly, g y (x, ) = 1/x for all x. ince, g x (, y) is not continues at y =, (g x ) y (, ) does not exist. imilarly, (g x ) y (, ) does not exist. Remark 1.9 The above example shows that a function can have partial derivatives at a point even if it does not have a limit at that point. EXAMPLE 1.17 Let z = f(x, y) := { xy(x 2 y 2 ), x 2 +y 2 (x, y) (, ), (x, y) = (, ). Note that f(x, ) = and f(, y) = for all x, y R. Therefore, f x (x, ) =, f y (, y) x, y R. and f xx (x, ) =, f yy (, y) x, y R. Now, by definition, (f x ) y (, ) = d dy f x(, y), (f y ) x (, ) = d y= dx f y(x, ). x= Note that and Thus, In particular, f( x, y) f(, y) f x (, y) = lim = y, x x f(x, y) f(x, ) f y (x, ) = lim = x. y y f xy (, y) = 1, f yx (x, ) = 1 x, y R. f xy (, ) = 1, f yx (, ) = 1. The above example shows that, in general, f xy need not be equal to f yx. However, under certain additional conditions they can be equal. Theorem 1.3 If f xy and fyx exist and are continuous in a neighbourhood of (x, y ), then f xy = f yx on.

18 Partial erivatives Partial Increments and Total Increment efinition 1.19 uppose f is a (real valued) function defined in a neigbourhood of a point (x, y), and let z = f(x, y). Then Partial increment of z with respect to x is x z := f(x + x, y) f(x, y). Partial increment of z with respect to y is Total increment of z is y z := f(x, y + y) f(x, y). z := f(x + x, y + y) f(x, y). Theorem 1.4 uppose f x and f y exist and are continuous in a neighbourhood 1 of a point (x, y). Then there exist functions ϕ and ψ defined in the neighbourhood 2 of (, ) such that z = f x (x, y) x + f y (x, y) y + φ( x, y) x + ψ( x, y) y for all (x, y) 1 and ( x, y) 2, where φ( x, y), ψ( x, y) as ( x, y) (, ). Proof. We may observe that z := f(x + x, y + y) f(x, y) = [f(x + x, y + y) f(x, y + y)] + [f(x, y + y) f(x, y)]. ince f has partial derivatives in a neighbourhood of (x, y), by mean value theorem, there exists ξ between x and x + x such that f(x + x, y + y) f(x, y + y) = f x (ξ, y + y), there exists η between y and y + y such that Thus, f(x, y + y) f(x, y) = f y (x, η). z = f x (ξ, y + y) x + f y (x, η) y. Further, since f x and f y are continuous at (x, y), f x (ξ, y + y) f x (x, y), f y (x, η) f y (x, y) as ( x, y) (, ), i.e., as ρ := ( x) 2 + ( y) 2. Thus, z = f x (x, y) x + f y (x, y) y + φ( x, y) x + ψ( x, y) y, where φ( x, y) and ψ( x, y) as ρ.

19 14 Functions of everal Variables Total ifferential, Gradient, ifferentiability uppose f is a function of two variables defined in a neighborhood of a point u := (x, y ). efinition 1.2 uppose function f has partial derivatives at (x, y ). Then the expression f x x + f y y is called the total differential. The total differential of z := f(x, y)f, and it is denoted by dz, and the infinitesimals x and y are called the differentials of the variables x and y and are denoted by dx and dy, respectively. Thus, dz = f x x + f y y or dz = f x dx + f y dy. The pair (f x, f y ) is called the gradient of f at u, denoted by (grad f)(u ) or ( f)(u ). Remark 1.1 In fact, the total differential of f at a point u is a function: ( x, y) f x x + f y y defined in a neighbourhood of (, ). By Theorem 1.4, z = dz. The above approximate equality can be used to approximate certain quantities. For instance, if an expression for f(x, y) is known, then f(x + x, y + y) = f(x, y) + dz. EXAMPLE 1.18 Volume of the material required to make a glass of inner radius r, inner height h and thickness k is: V = π(r + k) 2 (h + k) πr 2 h. Writing f(r, h) := πr h, the above expression can be written as r = k = h, f(r + r, h + h f(r, h). Thus, V = f(r + r, h + h f(r, h) = f r r + f h h = π(2rh r + r 2 h).

20 Partial erivatives x EXAMPLE 1.19 Let f(x, y) :=. Let us find an approximate (1 + y)(1 + z) express for f(x, y) when x and y are close to. This amounts to considering the approximate equality: f(x, y) = f(, ) + fx (, )x + f y (, )y = (1 + x y z). 2 efinition 1.21 The function f is said to be differentiable at (x, y ) if there exists a pair (α, β) of real numbers such that lim ρ 1 ρ {[f(x + x, y + y) f(x, y )] (α x + β y)} =, where ρ := ( x) 2 + ( y) 2. The pair (α, β) is called the derivative of f and is denoted by f (u ). From Theorem 1.4, the following result is obvious, thus providing a sufficient condition for differentiability at a point. Theorem 1.5 If f x and f y exist and are continuous in a neighbourhood of u = (x, y ), then f is differentiable at X, and f (u ) = ( f)(u ). Here is a necessary condition for differentiability. Theorem 1.6 uppose f is differentiable at u := (x, y ). Then, f is continuous, partial derivatives f x and f y exist at u, and f (u ) = ( f)(u ). Proof. Let z(u ) := f(x + x, y + y) f(x, y ). exists (α, β) R 2 such that By hypothesis, there Now, Hence, have z(u ) (α x + β y) lim =. ( ) ρ ρ [ ] z(u ) (α x + β y) z(u ) = ρ + (α x + β y). ρ lim z(u ) = so that f is continuous at u = (x, y ). From ( ) we also ρ f(x + x, y ) f(x, y ) α x lim = x x

21 16 Functions of everal Variables and f(x, y + y) f(x, y ) β y lim y x =. Hence f x and f y exist at u and we have f (u ) = (α, β) = (f x (u ), f y (u )). Remark 1.11 If a function f is either not continuous or if f x and f y does not exist at (x, y ), then By Theorem 1.6 we can conclude that f is not differentiable at (x, y ). uppose f x and f y exist at u = (x, y ). Then it follows that, f is differentiable at (x, y ) if and only if lim ρ 1 ρ {[f(x + x, y + y) f(x, y )] [f x (u ) x + f y (u ) y)]} =. and in that case the derivative is ( f)(x, y ) := (f x (x, y ), f y (u )) erivatives of Composition of Functions uppose z = F (u, v) where u and v are functions of (x, y), i.e., u = ϕ(x, y) and v = ψ(x, y) for some functions ϕ and ψ. Then z itself is a function of (x, y). Thus, Then we have z = F (ϕ(x, y), ψ(x, y)). x z = F (ϕ(x + x, y), ψ(x + x, y)) F (ϕ(x, y), ψ(x, y)). But, x u = ϕ(x + x, y) ϕ(x, y), x v = ψ(x + x, y) ψ(x, y). Hence, assuming all necessary conditions, we have where Thus, x z = F (u + x u, v + x v) F (u, v) = F u x u + F v x v + ϕ 1 x u + ϕ 2 x v, ϕ 1 ( x u, x v), ϕ 2 ( x u, x v) as ( x, y) (, ). x z x = F x u u x + F x v v x + ϕ x u 1 x + ϕ x v 2 x. Now taking limit as ( x, y) (, ), we have z x = F u u x + F v v x. Thus we have proved the following theorem.

22 Partial erivatives 17 Theorem 1.7 uppose ϕ and ψ are defined in a neighbourhood of a point (x, y ) and z = F (u, v) where u = ϕ(x, y), v = ψ(x, y) for (x, y). Assume that F u, F v exist and are continuous in a neighbourhood of (u, v ), where u = ϕ(x, y ), v = ψ(x, y ). Then for all (x, y) in a neighbourhood of (x, y ), we have z x = F u u x + F v v x. A special cases: (i) uppose f is a function of (x, y) in a nbd of a point (x, y ), and x and y are functions of another variable t with t [a, b]. Then z = f(x, y) is a function of t and we have dz dt = z t = f dx x dt + f dy y dt. (ii) uppose f is a function of (x, y) in a nbd of a point (x, y ), and y is a function x. Then z = f(x, y) is a function of x and we have dz dx = z x = f x + f y dy dx. EXAMPLE 1.2 Consider the function z = ln(u 2 + v 2 ) where u = e x=y2 v = x62 + y. Then and z x = z u u x + z v v x 2u = u 2 + v ex+y2 + 2x u 2 + v 2 ( ) = u 2 ue x+y2 + x + v and z y = z u u y + z v v y 2u = u 2 + v ex+y2 2y + 1 u 2 + v 1 ( ) = u 2 4ue x+y v EXAMPLE 1.21 Consider the function z = x 2 + y where y = sin x, < x < π. Then dz dx = z x + z dy cos x = 2x + y dx 2 y.

23 18 Functions of everal Variables Euler s Theorem: Theorem 1.8 uppose f is a homogeneous function of degree n in a domain, i.e., f(λx, λy) = λ n f(x, y) for all λ R, (x, y). Then Proof. We may write f(x, y) = f x f x + y f y = nf(x, y). ( x, x y ) ( = x n f 1, y ) = x n g(u), x x where u = y/x and g(u) = f(1, u). Then by Theorem 1.7, f x = nxn 1 g(u) + x n g (u)( y/x 2 ), f y = xn g (u)(1/x) = x n 1 g (u). From these expressions the conclusion of the theorem follows. 1.6 erivatives of Implicitly efined Functions efinition 1.22 A function y = f(x) is said to be implicitly defined on an interval J if there exists a function z = F (x, y) defined on a domain which contains the set {(x, ) : x J} such that F (x, f(x)) = x J. Theorem 1.9 uppose f is implicitly defined on an interval J of a point x by an equation F (x, y) = with y = f(x) for x J. Assume further that F x and F y are defined and are continuous in a nbd of a point (x, y ) where x J and y = f(x ). If F y at (x, y ), then F y and f (x) exists in a nbd J of x and f (x) = F x F y x J. Proof. Let x J and y = f(x ), y = f(x + x) f(x ). F (x, y ) = = F (x + x, y + y), we have Then, since = F = F x x + F y y + ϕ 1 x + ϕ 2 y, where ϕ 1 ( x, y), ϕ 2 ( x, y) as ( x, y). ince = F x + F y y x + ϕ 1 + ϕ 2 y x = (F x + ϕ 1 ) + (F y + ϕ 2 ) y x,

24 erivatives of Implicitly efined Functions 19 we have y x = F x + ϕ 1 F y + ϕ 2 F x F y as ( x, y). Thus f (x ) exists and f (x ) = F x /F y at (x, y ). ince F y at (x, y ) and F y is continuous in a nbd of (x, y ), we know that F y for all points in a nbd of (x, y ). In particular, there exists a nbd J of x such that F y (x, f(x)) for all x J. Hence, the result follows from the arguments in the above paragraph by replacing x by any point in J. uppose F is defined in a nbd of a point (x, y ). Question: oes there exists a function y = f(x) defined in a nbd J of x such that F (x, f(x)) = for all x J? The following theorem, known as the implicit function theorem prescribes certain conditions on F which guarantees affirmative answer to the above question. We omit its proof. Interested reader can see the proof in any of the books on Advanced Calculus, for example, the book, Advanced Calculus, by.v. Widder (Prentice-Hall of India, 1996). Theorem 1.1 (Implicit function Theorem) uppose F is defined in a nbd of a point (x, y ), and F x and F y exist and are continuous in. If F (x, y ) = and F y at (x, y ), then there exists a differentiable function y = f(x) defined in a nbd J of x such that (x, f(x)) and (i) F (x, f(x)) = for all x J, and (ii) f (x) = F x F y ((x, f(x)) for all x J. EXAMPLE 1.22 Let F (x, y) = x 2 + y 2 1 for (x, y) R 2. Clearly, F x and F y exist and are continuous in R 2. Also, F (x, y) = for every point on the circle := {(x, y) : x 2 + y 2 = 1}. Note that F y whenever (x, y) (1, ). In this case we have dy dx = x y at any point x with (x, y) and y. EXAMPLE 1.23 Let F (x, y) = e y e x +xy for (x, y) R 2. Clearly, F x := e x +y and F y := e y + x are continuous in R 2. Also, F y whenever e y x. In this case we have whenever e y x. dy dx = + y ex e y + x = ex y e y + x The above considerations can be extended to functions of more than two variables.

25 2 Functions of everal Variables o, let z = f(x, y) be defined implicitly by the equation F (x, y, z) =. Then under appropriate conditions on F and (x, y, z ), we have z x = F x F z, z y = F y F z. EXAMPLE 1.24 Let F (x, y, z) = x 2 + y 2 + z 2 R 2. Then z x = x z, z y = y z. EXAMPLE 1.25 Let F (x, y, z) = e z + x 2 y + z + 5. Then z x = F x F z = 2xy e 2 + 1, z y = F y = x2 F z e Level Curve and Level urface uppose z = u(x, y) is a function defined in a domain R 2, called a scalar field. Recall that the equations z = u(x, y) defines a surface. For c R, consider the surface defined by the equation z = c. One may visualize that the intersection of these two surface is a curve in the space. The projection of this curve onto the xy-plane is called a level curve. More precisely: efinition 1.23 if z = u(x, y) is a scalar field, then for each c R, the set of all (x, y) such that f(x, y) = c is called a level curve of u. efinition 1.24 If u is a function defined in a domain in R 3, then, for each c R, the set of all (x, y, z) such that u(x, y, z) = c is called a level surface of u. If Γ is a level curve of a function u, then it can be easily seen that v := (u y, u x ) is a vector in the direction of the tangent at the point (x, y). Indeed, if y = f(x) is implicitly defined by F (x, y) := u(x, y) c =, then the level curve of u is the graph of f, and its tangent is along the direction of (1, u x /u y ), i.e., along (u y, u x ) whenever u y, and if u y = then its direction is obviously along (, u x ). Thus, we see that u v = (u x, u y ) (u y, u x ) = u x u y u y u x =, so that the gradient u of u is perpendicular to the tangent line, and hence it is along the normal to the level curve.

26 irectional erivatives irectional erivatives uppose u is a function defined in a domain R 2. We would like to know the change in the values of u as (x, y) varies from (x, y ) along a particular direction s. Thus, taking s = ( x, y) and writing we have s = ( x) 2 + ( y) 2, u = u(x + x, y + y) u(x, y ) = u x x + u y y + ϕ 1 x + ϕ 2 y, where ϕ 1 and ϕ 2 as ( x, y) (, ). Hence, u s = u x x s + u y y s + ϕ x 1 s + ϕ y 2 s. Thus, if u has continuous partial derivatives at (x, y ), we have u lim s s = u x cos α + u y cos β, where α and β are the angles that s makes with the x-axis and y-axis respectively. Thus u u := lim s s s = u n s, where n s is the unit vector along s, that is, n s := s s. efinition 1.25 The quantity u s defined is called the directional derivative of u along s. Remark 1.12 We may observe by taking e 1 = (1, ) and e 2 = (, 1) that u = u e 1 x, u = u e 2 y. Remark 1.13 By the definition of the directional derivative, we have u s = u n s = u cos θ s, where θ s is the angle between the vectors u and s. Thus, u s u,

27 22 Functions of everal Variables and u s = u θ s =. Note that if s = u, then n s = u/ u so that u s = u n s = u. Thus, among all directional derivatives at a point, the one along s = u is the maximum. In a similar way, we can define directional derivatives of functions of three or more variables. Thus, if u = u(x, y, z) defined in a domain R 3, and if P = (x, y, z ), then the directional derivative of u along s := s 1 i + s 1 j + s 1 k is defined as u s = u n s = u cos θ s, where θ s is the angle between the vectors u and s. EXAMPLE 1.26 Let u(x, y) = xy. Then the directional derivative of u in the direction of s := (3, 4) at the point P = (1, 2) is given by u s := u n s at P, where u = (y, x) and n s = s/ s = 1 5 (3, 4). Hence, at P = (1, 2), u s = 1 5 (3y + 4x) P = 2. EXAMPLE 1.27 Let u(x, y) = x 2 + y 2. Then the directional derivative of u in the direction of s := (s 1, s 2 ) at the point P = (a, b) is given by u s := u n s at (a, b), where u = (2x, 2y) and n s = s/ s = (s 1, s 2 )/ s s2 2. Hence u s = (2xs 1 + 2ys 2 ) s s 2 2 P = 2 s s 2 2 (as 1 + bs 2 ). EXAMPLE 1.28 Let u(x, y, z 2 ) = x 2 + y 2 + z 2 and s = 2 s + j + 3 k. Then the directional derivative of u in the direction of s at the point P = (1, 1, 1) is: where Thus, u = (2x, 2y, 2z), so that at P = (1, 1, 1), u s u s = u n s, u = n s = s (2, 1, 3) (2, 1, 3) = =. s x + 2y + 6z 14 = (4 2) + (2 1) + (6 3) 14 =

28 Taylor s Formula 23 If s = u, u s (P ) = u (P ) = 2 x 2 + y 2 + z 2 (P ) = 2 3. Exercise 1.9 uppose f is differentiable at u := (x, y ). Then, prove that directional derivatives of f in any direction s exist at u, and 1.8 Taylor s Formula f (u ) = u s at u. Recall that if f is a function of one variable having continuous derivatives up to the order n + 1 in a nbd of a point x, then the Taylor s formula for f at any x in that nbd is f(x) = f(x ) + n k=1 f (k) (x ) k! (x x ) k + f (k+1) (ξ x ) (x x ) k+1 k! for some ξ x lying between x and x. Writing x = x x, the above formula can be written as n ( 1 f(x) = f(x ) + x d ) k ( 1 f (x ) + x d ) n+1 f(x + ξ x). k! dx (n + 1)! dx k=1 Here, we used the notation ( x d ) f := dx ( x df dx ), ( x d ) k ( f := x d ) ( x df ) k 1. dx dx dx We obtain similar formula for functions of two variables as well. For this purpose let us define ( x x + y ) ( f := x f ) f + y, y x y and for k = 2, 3,..., ( x x + y ) k ( f := x y x + y ) ( x y x + y ) k 1 f. y It can be seen that ( x x + y ) k f = y k r= ( ) ( ) k r ( ) k r C r ( x) r ( y) k r f. x y Note that, under the assumption that f has continuous partial derivatives f xy and f yx in a nbd of a point P, ( x x + y ) 2 f = ( x) 2 2 f y x x y 2 f x y + ( y)2 2 f y 2.

29 24 Functions of everal Variables Theorem 1.11 uppose f is a function of two variables having continuous partial derivatives up to the order n + 1 in a nbd of a point (x, y ). Then the Taylor s formula for any x in is given by f(x, y) = f(x, y ) + for some < ξ < (n + 1)! n k=1 ( 1 x k! ( x x + y y x + y ) k f(x, y ) y ) n+1 f(x + ξ x, y + ξ y). Proof. Let ϕ(t) = f(x + t x, y + t y). Then we have ϕ (k) (t) = ( x x + y ) k f(x + ξ x, y + ξ x). ( ) y Hence, using the Taylor s formula for functions of one variable we have ϕ(1) = ϕ() + n k=1 ϕ (k) () k! + ϕ(k+1) (ξ) k! for some ξ lying between and 1. formula. In view of ( ), this is exactly the required 1.9 Maxima and Minima efinition 1.26 uppose f : R and u. (i) The function f is said to have a maximum at u (or f attains local maximum at u ) if there exists a nbd G of u such that f(u) < f(u ) for all u G, u u, and we say that f attains a maximum at u. (ii) The function f is said to have a minimum at u (or f attains local minimum at u ) if there exits a nbd H of u such that f(u) > f(u ) for all u H, u u,, and we say that f attains a maximum at u. (iii) The function f is said to have extremum at u if f attains either maximum or minimum at u, and we say that f attains an extremum at u. Remark 1.14 (a) By the above definition, if a function f has a maximum (resp. minimum) at a point u, then there exists a nbd of u such that f can not attain maximum (resp. minimum) at any other point in that nbd. For example, the function f(x, y) = 1 defined on := {(x, y) : x 2 + y 2 < 1}, does not attain maximum or minimum at any point in, though for every u = (x, y ), f(x, y) f(x, y ) = 1 for every (x, y).

30 Maxima and Minima 25 (b) In the above we have defined maximum and minimum only locally, i.e., in a nbd of a point. In case there exists u such that f(x) < f(u ) (resp. f(x) > f(u )) for all u, then we say that f attains global maximum at u (global minimum at u ). Following theorem prescribes a necessary condition for a function to have a maximum or minimum at a point. Theorem 1.12 uppose f : R has a maximum or minimum at u, and suppose f x and f y exist at u. Then f x = = f y at u. Proof. uppose f : R has maximum at u = (x, y ). Then f(x + h, y ) < f(x, y ) for all h in a deleted nbd J of. Hence, the function ϕ : J R defined by ϕ(h) = f(x + h, y ) has a maximum at. Therefore, ϕ () =, i.e., f x (u ) =. imilarly, we can discuss the case of attaining minimum at u. efinition 1.27 Let f : R, and u. If f x and f y exist at u and f x = = f y at u, then u is called a critical point of f. A critical point of f at which f is neither a maximum nor a minimum is called a saddle point of f. EXAMPLE 1.29 Let f(x, y) = (x 1) 2 + (y 2) 2 1, (x, y) R 2. Then we see that f(x, y) > f(1, 2) = 1 for all (x, y) (1, 2). Thus, f attains minimum at (1, 2) R 2. EXAMPLE 1.3 Let f(x, y) = 1 2 sin(x2 + y 2 ), (x, y) R 2. Then we see that f(x, y) < f(, ) = 1 2 for all (x, y) (, ) in a nbd of (, ). Thus, f attains maximum at (1, 2) R 2. EXAMPLE 1.31 Let f(x, y) = x 2 y 2. Then f x = = f y at (, ) R 2, but, f attains neither maximum nor minimum at (, ); in fact, in every nbd of the origin, there are points at which f takes positive as well as negative values. Thus, (, ) is a critical point which is a saddle point of f. Now we prescribe a sufficient condition for a function to have a maximum or a minimum at a point. Theorem 1.13 uppose u := (x, y ) is a critical point of f : R, and f has continuous partial derivatives up to the order three in a nbd of u. Then we have the following: (i) f has a maximum at u, if (ii) f has a minimum at u if f xx f yy f 2 xy >, f xx < at u. f xx f yy f 2 xy >, f xx > at u.

31 26 Functions of everal Variables (iii) u is a saddle point of f if f xx f yy f 2 xy < at u. Remark 1.15 Look at the definition of a saddle point and the part (iii) in Theorem Part (iii) in Theorem 1.13 does not say that f xx f yy fxy 2 < is a necessary condition for u to be a saddle point of f; the condition is only sufficient. The point u can be a saddle point of f even when the condition f xx f yy fxy 2 < is not satisfied; but in that case one of the conditions f xx > and f xx < is to satisfied at u. Next we give another sufficient condition which involves eigenvalues of certain matrix. uppose A is a n n matrix with real entries. Then we say that λ R is an eigenvalue of A if det(a λi) =. It is known that if A is symmetric, then there are λ 1,..., λ n such that det(a λi) = (λ 1 λ)(λ 2 λ)... (λ n λ). A symmetric matrix A is said to be positive definite if all its eigenvalues are positive and it is said to be negative definite if all its eigenvalues are negative. Theorem 1.14 uppose u is a critical point of f : R, and f has continuous partial derivatives up to the order two in a nbd of u. Let [ ] fxx f H = xy at u. Then f yx f yy (i) f has a maximum at u, if H is negative definite, and (ii) f has a minimum at u if H is positive definite. (ii) u is a saddle point of f if H is neither positive definite nor negative definite, but det(h). EXAMPLE 1.32 Find the distance between the straight lines given by the equations (i) x 1 = y 1 2 = z and (ii) x 1 1 = y 1 = z 1. Note that an arbitrary point po the first line given by (λ + 1, 2λ, λ) whereas an arbitry point on the second line is given by (µ, µ, µ). Thus, we have to minimize the function f(λ, µ) := (λ + 1 µ) 2 + (2λ µ) 2 + (λ µ) 2. Note that f λ = 2(λ + 1 µ) + 4(2λ µ) + 2(λ µ) = 12λ 8µ + 2,

32 Maxima and Minima 27 f µ = 2(λ + 1 µ) 2(2λ µ) 2(λ µ) = 8λ + 6µ 2 = Hence, f λ =, f µ = give λ = 1/2, µ = 1. It can be seen that f λλ f µµ f 2 λµ >, f λλ > at (λ, µ) = (1/2, 1). Hence, the minimum distance is f(1/2, 1) = 1/ 2. EXAMPLE 1.33 Let f(x, y) = x 2 + y 2 xy + 3x 2y + 1. Then Hence and f x = 2x + y + 3, f y = 2y x 2. f x = = f y (x, y) = ( 4/3, 1/3) f xx f yy f 2 xy = 3 and f xx = 2 > at ( 4/3, 1/3). Hence, by the above theorem, f has a minimum at ( 4/3, 1/3). Remark 1.16 In Example 1.33, although f xx f yy fxy 2 = 3 and f xx = 2 at every (x, y) R 2, the function has only one extremum. EXAMPLE 1.34 Let us find the shortest distance from the origin to the plane x 2y 2z = 1 (by finding extrema of certain functions). Well, you already know that the shortest distance from the origin to the plane given by ax = by + cz = d is d / a 2 + b 2 + c 2. Thus, the answer required in the present example is 1/3. Let us find the same by finding the minimum of certain functions. Recall that the distance from the origin to any point (x, y, z) is given by x 2 + y 2 + z 2. Thus, the problem reduces to minimizing the function x 2 + y 2 + z 2 when (x, y, z) varies over the plane given by x 2y 2z = 1. ince x 2y 2z = 1 implies z = 1 2 (x 2y 1), it is enough to find the minimum value of f(x, y) := x 2 + y 2 + z 2 = x 2 + y (x 2y 1)2. Note that Thus, f x = 2x (x 2y 1) = 5 2 x y 1 2, f y = 2y (x 2y 1) = 4y x + 1. f x = & f y = x = 1 & y = 2 9 9, and in that case z = 2 9. ince the function f does not attain maximum at any point, the minimum of f is f( 1 9, 2 9 ) = 1 9 so that the required minimum distance is f( 1 9, 2 9 ) = 1 3.

33 28 Functions of everal Variables Exercise 1.1 how that the shortest distance from the origin to the plane given d by ax = by + cz = d is a, by finding minimum of certain function. 2 +b 2 +c2 Exercise 1.11 how that the rectangle having a maximum area for a fixed perimeter is a square Method of Lagrange Multipliers In Example 1.34 what we have found is the minimum of a function f(x, y, z) := x 2 + y 2 + z 2 (i) when (x, y, z) varies over the set of points such that ϕ(x, y, z) := x 2y 2z 1 =. (ii) The method we adopted was that, from the equation (ii), we expressed z as a function of (x, y), namely z = g(x, y) := 1 (x 2y 1) 2 and substituted the same into (i) to obtain a function of two variables, namely h(x, y) := f(x, y, g(x, y)) := x 2 + y 2 + x 2 + y (x 2y 1)2 4 and used the method of finding minimum of h. In general, suppose the problem is to find an extremum of a function f(x, y, z) (iii) when (x, y, z) varies over the set of points such that ϕ(x, y, z) =. (iv) If we are able to find a function z = g(x, y) such that then we could find extrema of ϕ(x, y, g(x, y)) =, h(x, y) := f(x, y, g(x, y)). But, if the function ϕ is not simple enough to find such a function g, then the above procedure cannot be adopted. Now we describe a procedure, called method of Lagrange multipliers, which does not involve such a function g explicitly. We shall consider the method of Lagrange multipliers in the case of two variables. The procedure can be extended to any finite number of variables.

34 Maxima and Minima 29 uppose we would like to find maximum or minimum of a function f which is defined in some open set R 2, subject to the condition ϕ(x, y) =, where ϕ is also defined in. uppose f attains maximum at a point P := (x, y ) as (x, y) varies over the set f := {(x, y) : ϕ(x, y) = }. Let us also assume that f x, f y, ϕ x, ϕ y exist in a nbd of P, and there exists a function y = g(x) defined in a nbd J of x such that ϕ(x, g(x)) = for all x J. Then u = f(x, y) is a function of a single variable x J. Thus, u attains maximum at P so that derivative of u w.r.t. x is at x. Thus, we have the following necessary conditions: du dx := f x + f y y =, ϕ x + ϕ y y = at P. Hence, we must have i.e, (f x + f y y ) + λ(ϕ x + ϕ y y ) = at P λ R, (f x + λϕ x ) + (f y + λϕ y )y = at u λ R, Choosing λ such that f y + λ ϕ y = at (x, y ), we obtain ϕ =, f x + λϕ x =, f y + λϕ y = for (λ, x, y) = (λ, x, y ). ( ) Note that, writing the conditions in ( ) is same as F (x, y, λ) := f(x, y) + λϕ(x, y) ϕ =, F x = = F y at (λ, x, y ). The parameter λ above is called the Lagrange multiplier, and the method using Lagrange multiplier is the procedure of finding (λ, x, y) such that ϕ =, F x = = F y so that the required point at which f attains an extremum in f is one among these points. EXAMPLE 1.35 Among all rectangles with a given perimeter l, let us find the one having maximum area. Thus, the problem is to find the point (x, y ) at which the function f(x, y) := xy attains maximum subject to the constraint ϕ(x, y) := 2(x + y) l =. We consider the equation ϕ = 2(x + y) l =, f x + λϕ x := y + 2λ =, f y + λϕ y = x + 2λ =. Thus, so that x = y = l/4. x = 2λ = y, l = 2(x + y) = 4x

35 3 Functions of everal Variables EXAMPLE 1.36 We show that among all rectangular parallelepiped inscribed in a given sphere, cube has the maximum volume. To see this, let x, y, z be the sides of the parallelepiped. Clearly, we must have x 2 + y 2 + z 2 = d 2, where d is the diameter of the sphere. o, we must find the maximum of the function f(x, y, z) := xyz subject to the condition ϕ(x, y, z) := x 2 + y 2 + z 2 d 2. Hence, the equations to be solved are: ϕ(x, y, z) := x 2 + y 2 + z 2 d 2 = From the above equations, it follows that f x + λϕ x := yz + λ(2x) = f y + λϕ y := xz + λ(2y) = f x + λϕ z := xy + λ(2z) =. x(f x + λϕ y ) + y(f y + λϕ y ) + z(f z + λϕ z ) =, i.e., Thus, Hence, 3xyz + 2λ(x 2 + y 2 + z 2 ) =. 3xyz + 2λd 2 =. = yz + 2λ2x = yz 3 ( d 2 x2 yz = yz 1 3 d 2 x2). Thus, x = d/ 3. imilarly, y = d/ 3 and z = d/ 3. EXAMPLE 1.37 Let us find the parelleloeiped of maximum volume with a given surface area: The function to be maximized is f(x, y, z) = xyz subject to the condition 2(xy + yz + zx) = A. Thus, ϕ(x, y, z) = 2(xy + yz + zx) = A. Note that f x + λϕ x = yz + 2λ(y + z) =, f y + λϕ y = zx + 2λ(z + x) =, f z + λϕ z = xy + 2λ(x + y) =. Multiplying (i), (ii) and (iii) by x, y and z respectively and adding we get (i) (ii) (iii) 3xyz + 2λA =.

36 Maxima and Minima 31 Thus, λ = 3xyz/2A. Hence, from (i), i.e., yz 3 xyz(y + z) =, i.e., 3x(y + z) = A, i.e., 3(xy + xz) = A, A 3(A 2yz) = 2A,, i.e., 6yz = A,, i.e., yz = A/6. imilarly, zx = A/6 and xy = A/6. Thus, so that z = A/6y and hence, x = A/6z = y. Hence, x 2 = A/6 and x = A/6. imilarly, y = A/6 and z = A/6. EXAMPLE 1.38 uppose a wire of length l is cut into three pieces and are bent them to form a circle, a square, and an equilateral triangle. Let us find the length of these pieces so that that the total areas inscribed by these figures is minimum: The function to be minimized is subject to the constrained f(x, y, z) := πx 2 + y z2 ϕ(x, y, z) := 2πx + 4y + 3z l =. Now, we may apply Lagrange multiplier method. The equations to be solved are: Thus, ϕ(x, y, z) := 2πx + 4y + 3z l = f x + λϕ x := 2πx + λ(2π) = f y + λϕ y := 2y + λ(4) = 3 f z + λϕ z := z + λ(3) =. 2 x = λ, y = 2λ, z = λ 6 3 ( ) and ( ϕ(x, y, z) := 2πx + 4y + 3z l = 2π( λ) + 4( 2λ) + 3 λ 6 ) l = 3 Hence [ ( 6 )] λ 2π = l. ( ) From ( ), we get value of λ, and then obtain the values of x, y, z, from the previous three equations: f(x, y, z) = πx 2 + y z2 = πλ 2 + 4λ λ 2 = (π )λ 2.