Definitions and claims functions of several variables

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1 Definitions and claims functions of several variables In the Euclidian space I n of all real n-dimensional vectors x = (x 1, x,..., x n ) the following are defined: x + y = (x 1 + y 1, x + y,..., x n + y n ) - vector addition cx = (cx 1, cx,..., cx n ) - multiplication of a vector by a scalar x y =< x, y >:= x 1 y 1 + x y + + x n y n - dot product (inner product, scalar product) x = x x = x x n - norm (length) of a vector d(x, y) = x y = (x 1 y 1 ) +... (x n y n ) - distance of x from y If θ is the angle between the nonzero vectors x and y then x y = x y cos θ. Definition. Given a point (vector) a I n, we defined the ball with center a and radius ε (the ε-neighbourhood of a) as B(a, ε) = {x I n : x a < ε}. Definition. Let D I n, a function f : D I is a function of several variables. We usually indicate f as f(x). If n =, f(x, y) is a function of two variables, if n = 3, f(x, y, z) is a function of three variables. D is the domain of f and {f(x) : x D} is the range of f. Example. f(x, y) = x+y+1 x 1, D = {(x, y) : x + y + 1 0, x 1} is the half plane above the line y = x 1 without the points on the line x = 1. Example. f(x, y, z) = xy ln z, D I 3 is the half space z > 0. Definition. If f is a function of two variables with domain D, the graph of f is the set {(x, y, z) I 3 : z = f(x, y), (x, y) D}. Example. f(x, y) = 9 x y, D is the disk with center (0, 0) and radius 3, the graph of f is the top half of the sphere with center at the origin and radius 3. Definition. The level curves of a function f of two variables are the curves with equation f(x, y) = k, where k is a constant. Example. f(x, y) = 4x + y, elliptic paraboloid, its level curves x + y = k are ellipses. emark. For functions of three variables we talk of level surfaces: for f(x, y, z) = x +y +z the level surfaces are spheres. Definitions and claims functions of several variables - Limits and continuity. Definition. Let f be a function of several variables defined on a neighbourhood of a I n (a ball with center at a) except possibly at a. Then we say that the it of f(x) as x approaches a is L f(x) = L x a if for every ε > 0 there exists a corresponding δ > 0 such that f(x) L < ε whenever 0 < x a < δ. emark. indicates the absolute value, i.e. f(x) L is the distance of f(x) from L in I, while indicates the norm, i.e. x a the distance of x from a in I n. Example. Discuss x y (x,y) (0,0) x +y. Let s approach (0, 0) along the axis: x axis, y = 0 f(x, 0) = x x = 1 so f(x, y) 1 along x axis; y axis, x = 0 f(x, 0) = y y = 1 so f(x, y) 1 along y axis; thus, the it does not exist. Fact. If a function approaches different its along two different paths, then the it does not exist (DNE). Example. (x,y) (0,0) (x,y) (0,0) xy (x,y) (0,0) x +y DNE (zero along the axis, 1/ along y = x). xy x +y 4 DNE (zero along the axis and along every line y = mx, 1/ along x = y ). 8x y x +y = 0, ε-δ-proof with δ = ε/8. (x,y,z) (0,0,0) ln(x + y + z ) = 1

2 Definition. Let f be a function of several variables defined on a ball B a, then f is continuous at a if f(x) = f(a). x a This means that three conditions must be satisfied: 1) the function is defined at a; ) the it exists; 3) the it equals the value of the function at a. A function f is continuous on a set D if it is continuous at every point in D. Example. f(x, y) = 8x y { 8x y x +y is not continuous at (0, 0) because it is not defined there; x f(x, y) = +y (x, y) (0, 0) is not continuous at (0, 0) because the it is different from f(0, 0); { 1 (x, y) = (0, 0) 8x y x f(x, y) = +y (x, y) (0, 0) is continuous everywhere. 0 (x, y) = (0, 0) emark. All Theorems about its and continuity studied for functions of one variable are extendable to functions of several variables: uniqueness of it, it and continuity of sum, difference, product, quotient, composition of functions. In particular, if f is a continuous function defined on a closed bounded set D of I n, then f obtains its maximum and minimum value in D. Example. Find domain, discuss continuity and graph the function f(x, y) = ln(x + y 1). Definitions and claims functions of several variables - Partial derivatives. Definition. If f is a function of two variables x and y, then its partial derivative at (x 0, y 0 ) with respect to x is f f(x 0 + h, y 0 ) f(x 0, y 0 ) x = f x (x 0, y 0 ) = D x f := (x0,y h 0 0) h if the it exists, its partial derivative at (x 0, y 0 ) with respect to y is f f(x 0, y 0 + h) f(x 0, y 0 ) y = f y (x 0, y 0 ) = D y f := (x0,y h 0 0) h if the it exists. If we let (x 0, y 0 ) vary, f x, f y are the functions partial derivatives defined by f(x + h, y) f(x, y) f(x, y + h) f(x, y) f x (x, y) =, f y (x, y) = h 0 h h 0 h Example. f(x, y) = x 3 + xy sin xy + 6, f x = 3x + y y cos xy, f y = xy x cos xy. emark. Geometrically, if the graph of f is the surface S, then the partial derivative f x (x 0, y 0 ) represents the slope of the tangent line at the point P (x 0, y 0, f(x 0, y 0 )) S to the curve resulting from the intersection of the surface S with the plane y = y 0. Analogues interpretation for f y (x 0, y 0 ): slope of tangent line at P to the curve intersection of S and x = x 0. Example. f(x, y) = y y+e yx : find f x, f y, f x (1,1). Definition. For functions of more variables, f(x 1,..., x n ), f f(x 1,..., x i + h,..., x n ) f(x 1,..., x i,..., x n ) = x i h 0 h Example. f(x, y, z) = e xy ln z, find all partial derivatives. Definition. Since partial derivatives are themselves functions of several variables, we define their partial derivatives ((f x ) x, (f x ) y, (f y ) y, (f y ) x ) as the second order partial derivatives of f. Moreover we use the notation (f x ) x = f xx = f x, (f x ) y = f xy = f y x, (f y) y = f yy = f y, (f y ) x = f yx = f x y. Similar definition and notation are introduced for functions of three or more variables, and for higher order derivatives. Example. f(x, y) = x 3 + e x cos y + xy y, find all second order partial derivatives. g(x, y, z) = xy z z sin(x yz), find f zxy. Theorem. Suppose f : I n I is defined on a neighborhood of a point and has continuous second order partial derivatives at the point, then f xix j = f xjx i, where 1 i, j n (mixed partial derivatives are the same). emark. A function may have partial derivatives at a point without being there continuous.

3 { 0 xy 0 Example. f(x, y) = is differentiable but not continuous at (0, 0). 1 xy = 0 Definition. A differential equation involving partial derivatives of a function of several variables is called a partial differential equation. f x + f y = 0 is called Laplace equation, its solutions are harmonic functions. Example. Show that f(x, y) = e x sin y solves the Laplace equation. Consider a function of two variables z = f(x, y). If x and y are given increments x, y, then z has a corresponding increment z = f(x + x, y + y) f(x, y). Theorem. Suppose f x, and f y exist on a region containing (x 0, y 0 ) and (x 0 + x, y 0 + y). Suppose that f x, and f y are continuous at (x 0, y 0 ), then where ε 1 0, ε 0 as ( x, y) (0, 0). Proof. (Use Lagrange ) z = f x (x 0, y 0 ) x + f y (x 0, y 0 ) y + ε 1 x + ε y Definition. If z = f(x, y), then f is differentiable at (x 0, y 0 ) if z can be written in the form z = f x (x 0, y 0 ) x + f y (x 0, y 0 ) y + ε 1 x + ε y where ε 1 0, ε 0 as ( x, y) (0, 0). f is differentiable on a region if it is differentiable at every point of the region. Corollary. If f x and f y are continuous on a region, then f is differentiable on. Theorem. If f is differentiable at a point, then f is continuous at the point. Definition. Given a function f(x, y) differentiable at (x 0, y 0 ), we define the linearization of f at (x 0, y 0 ) as the function l(x, y) = f(x 0, y 0 ) + f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ) l(x, y) is the linear approximation of f at (x 0, y 0 ). Suppose that f has continuous first and second order partial derivatives in a rectangular region centered at (x 0, y 0 ), and M is a upper bound for the values of f xx, f yy, f xy on, then the error E(x, y) we make by replacing the function f with its linearization on can be estimated by E(x, y) = 1 M( x x 0 + y y 0 ). Definition. z = l(x, y) is the equation of the tangent plane to the surface z = f(x, y) at (x 0, y 0, f(x 0, y 0 )). If z 0 = f(x 0, y 0 ) the equation of the tangent plane is indeed z z 0 = f x (x 0, y 0 )(x x 0 ) + f y (x 0, y 0 )(y y 0 ). Example. Find equation of tangent plane to z = e x ln y at (3, 1, 0). The definition of linearization can be extended to functions of more then three variables (and in general for function from I n to I m, we will reconsider this definition after introducing the concepts of gradient and total differential). Example. Given f(x, y, z) = x xy + 3 sin z, find its linearization at (x 0, y 0, z 0 ) = (, 1, 0). Use the result to approximate the value of f at (, 1, 0.01) and at (.1, 1, 0). Definitions and claims functions of several variables - Chain rule. Implicit differentiation. Theorem. Suppose z = f(x, y) is a differentiable function, and suppose x = g(t) and y = h(t) with both g and h differentiable functions of t. Then f is a differentiable function of t and df dt = dz dt = f dx x dt + f dy y dt, z x y t t It is convenient to draw a tree diagram to visualize the dependent, intermediate and independent variables. Example. Given z = x y + 3xy 4, x = e t, y = sin t, find dz dz dt and dt t=0. 3

4 Theorem. The above theorem can be extended to function f of n variables x 1, x,..., x n where each x j is a function of m variables t 1,..., t m, then if all partial derivatives xj t i exist, we have for every i = 1,..., m. f t i = n j=1 f x j x j t i Example. Evaluate dw dt if w = f(x, y, z) = xy + z and x = cos t, y = sin t, z = t. (The function f is consider over a path, indeed x = cos t, y = sin t, z = t are the parametric equations of a helix in I 3 ). In this case using the tree diagram w dw x y z we get dt = w dx x dt + w dy y dt + w dz z dt. t t t Example. Given z = e x sin y, x = st, y = s t, evaluate z t and z s. From z x y s t s t we get z t = z x x t + z y y t, z s = z x x s + z y y s. Example. Given w = x + y + z, x = r s, y = r ln s, z = r, find w r w x y z. s r s r s r w and s using the tree diagram Theorem. Implicit Function Theorem. F (x, y) = 0 defines implicitly y as a function of x near a point (a, b) if F (a, b) = 0, F y (a, b) 0 and F x and F y are continuous on a disk containing (a, b). Theorem. If F (x, y) = 0 defines implicitly y as a function of x then the derivative of y with respect to x can be evaluated by dy dx = F x. F y Proof. (Use chain rule.) Example. Given F (x, y) = x + sin y y use implicit differentiation to find dy dx near (0, 0). (Prove first that F (x, y) = 0 defines implicitly y as a function of x near (0, 0).) 5. Directional derivative. Gradient. If z = f(x, y), the partial derivatives f x, f y represent the rate of change of f in the direction of the x and y axis. We may wish to find the rate of change of z in any direction. emember that given any vector v its direction is determine by the unit vector u = v v. Definition. The directional derivative of f at (x 0, y 0 ) in the direction of the unit vector u = (a, b) is D u f(x 0, y 0 ) = h 0 f(x 0 + ha, y 0 + hb) f(x 0, y 0 ) h if the it exists. Theorem. If f is a differentiable function of x and y, then f has directional derivative in the direction of any unit vector u = (a, b) and D u f(x, y) = f x (x, y)a + f y (x, y)b. Proof. (Use chain rule.) Example. Find the directional derivative of f(x, y) = x 3 y 4 at P (6, 1) in the direction of v =<, 5 >= i+5j. Definition. Given f(x, y) we call gradient of f the vector function f =< f x, f y >= f x i + f y j. 4

5 Example. Given f(x, y) = cos x e xy, evaluate f and f (0, 1). emark. D u f(x, y) = f u ( inner product of two vectors). emark. Due to the fact that v u = v u cos θ, where θ is the angle between v and u, we have that D u is maximal when u and f are parallel. Thus: f(x, y) increases most rapidly in the direction of f (at every point). Any direction u to f is a direction of no change in f, so f is to level curves. The direction of f is the direction of minimal change of rate of f. emark. In three dimensions: f =< f x, f y, f z >, f is (still) the direction of maximal rate of increase, f (if 0) is orthogonal to level surfaces, so it is to the tangent plane to the level surface. Example. Find the tangent plane to the circular paraboloid x + y + z 9 = 0 at P (1,, 4). (Notice that if f(x, y, z) = x + y + z 9 the paraboloid is the level surface f = 0). For the general case of a function f : I n I, the definition and theorems concerning Directional derivatives are the following. Definition. The directional derivative of a function f : I n I in the direction of a unit vector u at the point x is f(x + hu) f(x) D u f(x) =. h 0 h Theorem. If x is a point such that the gradient of f, f =< f x 1,..., f at x, then D u f(x) = f(x) u. Definition. The linearization of a function f(x) at a point a is: f(x) f(a) + f a (x a). x n > is a continuous vector function Definitions and claims Vector functions. The total derivative as a linear map. Definition. A function f : I n I m is called a vector valued function. Such a function is given by m real-valued component functions, f(x) =< f 1 (x 1,..., x n ),..., f m (x 1,..., x n ) >, with x =< x 1,..., x n >. The partial derivatives of all component functions (if they exist) can be organized in an m n matrix, the Jacobian matrix J of f, or J f, as follows: J = f 1 x 1. f m x 1 f 1 x n.... Definition. Let U I n be an open subset. Then a function f : U I m is said to be (totally) differentiable at a point p U, if there exists a linear map d f p : I n I m (also denoted D p f or Df(p)) such that f m x n. f(x) f(p) d f p (x p) = 0. x p x p The linear map d f p is called the (total) derivative or (total) differential of f at p. A function is (totally) differentiable if its total derivative exists at every point in its domain. emark. Note that f is differentiable if and only if each of its components f i : U I is differentiable. For this it is necessary, but not sufficient, that the partial derivatives of each function f j exist. However, if these partial derivatives exist and are continuous, then f is differentiable and its differential at any point is the linear map determined by the Jacobian matrix of partial derivatives at that point. Thus in terms of linearization we can say that f(x) f(p) + J f p (x p). emark. We must observe that the above definition generalizes the one given in our previous class for m = 1; for functions f : I n I the Jacobian matrix corresponds to the gradient f of f. Example. Linearize f : I I f(x, y) =< x + y, 3x + y > around the point (1, 1). Definitions and claims functions of several variables - Maximum and minimum values. Definition. Given f(x, y) defined on a region, we say that f has a local maximum at (a, b) (and f(a, b) is the local maximum value) if f(x, y) f(a, b) for every point (x, y) in a certain disk with center (a, b), 5

6 we say that f has a local minimum at (a, b) (and f(a, b) is the local minimum value) if f(x, y) f(a, b) for every point (x, y) in a certain disk with center (a, b). If the above inequalities hold for every point in the domain of f, we then say that f has absolute maximum or absolute minimum at (a, b). Theorem. If f has local minimum or local maximum at (a, b), then f x (a, b) = f y (a, b) = 0. Proof. Definition. A point (a, b) such that f x (a, b) = 0 and f y (a, b) = 0, or one of the partial derivatives does not exist, is called a critical point. Definition. f(x, y) has a saddle point at a critical point (a, b) if for every disk D such that (a, b) D there are points (x, y) D such that f(x, y) f(a, b) and points (x, y) D such that f(x, y) f(a, b). Example. f(x, y) = x + y has absolute minimum at (0, 0). Example. For z = y x, (0, 0) is a saddle point. (Same for z = xy, z = y y 4 x, z = xy(x y ) (x +y ), use maple to graph these functions). Theorem. If f(x, y) is continuous and has continuous second partial derivatives on a disk D containing (a, b), suppose that f x (a, b) = f y (a, b) = 0. Consider then the value (a, b) = f xx f xy the determinant of the Hessian matrix, f yx = f xy f yy i.e. the Jacobian of the gradient of f at (a, b) (a,b) If > 0 and f xx (a, b) > 0, then f(a, b) is a local minimum. If > 0 and f xx (a, b) < 0, then f(a, b) is a local maximum. If < 0, then f(a, b) is a saddle. Example. Find local extrema (maximum, minimum or saddle points) of f(x, y) = xy x y x y + 4. If a function is continuous on a closed bounded set, then it obtains its absolute maximum and absolute minimum values. To find the absolute maximum and minimum on a closed bounded set D we then follows the following Step 1. Find critical points of f inside D and evaluate the function at each point. Step. Find extreme points of f on the boundary of D, and evaluate the function at those points. Step 3. Compare all the values from 1. and., the largest is the absolute maximum value of f on D and the smallest is the absolute minimum of f on D. Example. Find absolute maximum and minimum of f(x, y) = + x + y x y on the triangle with vertices at (0, 9), (0, 0) and (9, 0) (sides included). Definitions and claims functions of several variables - Lagrange multipliers. The method of Lagrange multipliers provides a strategy for finding the maximum and minimum of a function subject to constraints. We consider a problem of the type Maximize f(x) subject to g(x) = k, where x I or x I 3. Assume that these extreme values exist. The method of Lagrange multipliers then says that we need to find all points x and λ I such that f(x) = λ g(x) and g(x) = k and evaluate the function f at all such points x. The biggest value is the maximum of f, the smallest the minimum. Proof: (For x I.) Example. Find maximum and minimum value of f(x, y) = xy on the curve x 8 + y = 1. Definitions and claims Multiple integrals. Double integrals. Definition. Consider a function f(x, y) defined over a rectangle = [a 1, b 1 ] [a, b ]. Consider a partition a 1 = x 1 < x < < x m = b 1, of [a 1, b 1 ] and a partition a = y 1 < y < < y n = b, of [a, b ], then the set of all mn subrectangles ij = [x i 1, x i ] [y j 1, y j ] forms a partition of. We call norm of this partition of the length δ of the longest diagonal of all subrectangles ij. Also the area of the ij-subrectangle is A ij = (x i x i 1 )(y j y j 1 ). Chose a point (x ij, y ij ) inside each ij, we now can define the iemann sum m i=1 j=1 n f(x ij, yij) A ij. 6

7 The function f is said to be iemann integrable over, and the double integral of f over is f(x, y) da = m δ 0 i=1 j=1 n f(x ij, yij) A ij if the it exists. Theorem. If f(x, y) 0, and f is integrable over a rectangle, then the volume of the solid that lies under z = f(x, y), above the rectangle is V = f(x, y) da. emark. If we want to approximate the volume, we can consider a finite partition of and evaluate a finite sum as in the definition of double integral. Example. Use a finite partition of = [0, ] [0, 1] to approximate the volume of the solid lying under z = 4 x y and above. Theorem. (Fubini.) If f(x, y) is integrable over the rectangle = [a, b] [c, d] then Example. f(x, y) da = b a ( d c ) f(x, y) dy dx = d c ( b Evaluate (4 x y) da using both orders of integration. b d a emark. c h(x)g(y) dydx = b a h(x) dy d g(y) dy. c Theorem. (Properties of double integral.) kf da = k f da k I, f ± g da = f da ± A f da, f da 0 if f 0, f da g da if f g, f da = 1 f da + f da if 1 and 1 =, f da f da. a ) f(x, y) dx dy. If we need to integrate f(x, y) over a non-rectangular bounded region D, we close the region inside a rectangle and we define the function { f(x, y) (x, y) D g(x, y) = 0 (x, y) \ D then we define D f(x, y) da = g(x, y) da. Theorem. If the region D can be described as D = {(x, y) : a x b, g 1 (x) y g (x)}, then f(x, y) da = b a ( g(x) g 1(x) ) f(x, y) dy dx. If the region D can be described as D = {(x, y) : c y d, h 1 (y) y h (y)}, then f(x, y) da = d c ( h(y) h 1(y) ) f(x, y) dx dy. Example. Integrate in two different possible ways f(x, y) = 3 x y on the triangle D with vertices at (0, 0), (1, 0), (1, 1). emark. The area of a region A I is calculated as a double integral: Area(A)= A da. emark. The area of the surface with equation z = f(x, y), (x, y) D, where f x and f y are continuous is Area(S) = fx (x, y) D + f y (x, y) + 1 da. Definitions and claims Integral in polar coordinates. Change of variables. emark. The description of certain type of regions in I is easier with the use of polar coordinates. For example, the closed disk with center at the origin and radius one is described in polar coordinates by D = {(r, θ) : 0 r 1, 0 θ π}, thus is a rectangle, while in cartesian coordinates by D = {(x, y) : 1 x 1, 1 x y 1 x }. emember that the relation between cartesian and polar coordinates is x = r cos θ, y = r sin θ. Fact. If f is integrable on a region D, then D f(r, θ) da = f(r, θ) r dr dθ. D Example. Evaluate D (3x + 4y ) da, where D = {(x, y) : y 0, 1 x + y 4} with the use of polar coordinates. 7

8 Theorem. Suppose T : I I is a C 1 -transformation given by x = g(u, v), y = h(u, v), that maps a region S in the uv-plane onto a region D in the xy-plane then f(x, y) dx dy = f(g(u, v), h(u, v)) det J(u, v) du dv D S where det J(u, v) is the absolute value of the determinant of the Jacobian of the transformation, i.e. det J(u, v) = x y u v x y v u. emark. Changing an integral from cartesian to polar coordinates we use the transformation x = r cos θ, y = r sin θ and in this case D f(x, y) dx dy = f(r cos θ, r sin θ)r dr dθ, being det J(r, θ) = r. S Example. Use the transformation x + y = u, x y = v to evaluate D e x+y x y dxdy where D I is the region with vertices (0, 1), (0, ), (1, 0), (, 0). 8

Review guide for midterm 2 in Math 233 March 30, 2009

Review guide for midterm 2 in Math 2 March, 29 Midterm 2 covers material that begins approximately with the definition of partial derivatives in Chapter 4. and ends approximately with methods for calculating