Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables.

Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector. Directional derivative of functions of two variables. Remark: The directional derivative generalizes the partial derivatives to an direction. Definition The directional derivative of the function f : D R 2 R at the point P 0 = ( 0, 0 ) D in the direction of a unit vector u = u, u is given b ( Du f ) P0 = lim t 0 if the limit eists. 1[ f (0 + u t, 0 + u t) f ( 0, 0 ) ], t Notation: The directional derivative is also denoted as ( df ). dt u,p0

00 00 00 01 00 00 01 01 01 11 11 00 00 01 01 01 00 Directional derivatives generalize partial derivatives. Eample The partial derivatives f and f are particular cases of directional derivatives ( D u f ) P0 = lim t 0 1 t [ f (0 + u t, 0 + u t) f ( 0, 0 ) ] : u = 1, 0 = i, then ( D i f ) P0 = f ( 0, 0 ). u = 0, 1 = j, then ( D j f ) P0 = f ( 0, 0 ). z f (, ) f (, ) 0 f ( 0, 0 ) z f (,) f (, ) 0 0 ( D f ) u P0 f (, ) 0 00 00 00 00 00 11 11 11 11 11 11 00 00 00 11 11 11 11 11 00 00 00 11 11 11 11 11 i ( 0, ) 0 01 j 00 0000 11 11 00 00 1111 0011 0011 11 P = (, ) 0 0 0 01 11 u u =1 Directional derivative of functions of two variables. Remark: The condition u = 1 implies that the parameter t in the line r(t) = 0, 0 + u t is the distance between the points ((t), (t)) = ( 0 + u t, 0 + u t) and ( 0, 0 ). Proof. that is, d = t. d = 0, 0, = u t, u t, = t u, Remark: The directional derivative of f (, ) at P 0 = ( 0, 0 ) along u, denoted as ( D u f ), is the pointwise rate of change of f P0 with respect to the distance along the line parallel to u passing through ( 0, 0 ).

Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector. Directional derivative and partial derivatives. z f (,) Remark: The directional derivative ( D u f ) is the P0 derivative of f along the line r(t) = 0, 0 + u t. 00 00 1111 00 00 00 11 11 11 00 0000 00 00 00 00 00 11 11 11 11 11 11 00 00 11 11 11 11 01 P = (, ) 0 0 0 ( D f ) u 00 0 00 00 00 00 11 11 11 11 1 11 000 00 11 111 01 01 Theorem If the function f : D R 2 R is differentiable at P 0 = ( 0, 0 ) and u = u, u is a unit vector, then u P0 u =1 ( Du f ) P0 = f ( 0, 0 ) u + f ( 0, 0 ) u.

Directional derivative and partial derivatives. Proof. The line r(t) = 0, 0 + u, u t has parametric equations: (t) = 0 + u t and (t) = 0 + u t; Denote f evaluated along the line as ˆf (t) = f ((t), (t)). Now, on the one hand, ˆf (0) = ( D u f ) P0, since ˆf 1 ] (0) = lim [ˆf (t) ˆf (0), t 0 t 1[ = lim f (0 + u t, 0 + u t) f ( 0, 0 ) ], t 0 t = D u f ( 0, 0 ). On the other hand, the chain rule implies: ˆf (0) = f ( 0, 0 ) (0) + f ( 0, 0 ) (0). Therefore, ( D u f ) P0 = f ( 0, 0 ) u + f ( 0, 0 ) u. Directional derivative and partial derivatives. Eample Compute the directional derivative of f (, ) = sin( + 3) at the point P 0 = (4, 3) in the direction of vector v = 1, 2. Solution: We need to find a unit vector in the direction of v. Such vector is u = v u = 1 1, 2. v 5 We now use the formula ( D u f ) P0 = f ( 0, 0 ) u + f ( 0, 0 ) u. That is, ( D u f ) P0 = cos( 0 + 3 0 )(1/ 5) + 3 cos( 0 + 3 0 )(2/ 5). Equivalentl, ( D u f ) P0 = (7/ 5) cos( 0 + 3 0 ). Then, ( D u f ) P0 = (7/ 5) cos(10).

Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector. Directional derivative of functions of three variables. Definition The directional derivative of the function f : D R 3 R at the point P 0 = ( 0, 0, z 0 ) D in the direction of a unit vector u = u, u, u z is given b ( Du f ) P0 = lim t 0 if the limit eists. 1[ f (0 + u t, 0 + u t, z 0 + u z t) f ( 0, 0, z 0 ) ], t Theorem If the function f : D R 3 R is differentiable at P 0 = ( 0, 0, z 0 ) and u = u, u, u z is a unit vector, then ( Du f ) P0 = f ( 0, 0, z 0 ) u + f ( 0, 0, z 0 ) u + f z ( 0, 0, z 0 ) u z.

Directional derivative of functions of three variables. Eample Find ( D u f ) for f (,, z) = 2 + 2 2 + 3z 2 at the point P0 P 0 = (3, 2, 1) along the direction given b v = 2, 1, 1. Solution: We first find a unit vector along v, u = v v u = 1 6 2, 1, 1. Then, ( D u f ) is given b ( D u f ) = (2)u + (4)u + (6z)u z. We conclude, ( D u f ) P0 = (6) 2 6 + (8) 1 6 + (6) 1 6, that is, ( D u f ) P0 = 26 6. Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector.

The gradient vector and directional derivatives. Remark: The directional derivative of a function can be written in terms of a dot product. In the case of 2 variable functions: D u f = f u + f u D u f = ( f ) u, with f = f, f. In the case of 3 variable functions: D u f = f u + f u + f z u z, D u f = ( f ) u, with f = f, f, f z. The gradient vector and directional derivatives. Definition The gradient vector of a differentiable function f : D R 2 R at an point (, ) D is the vector f = f, f. The gradient vector of a differentiable function f : D R 3 R at an point (,, z) D is the vector f = f, f, f z. Notation: For two variable functions: f = f i + f j. For two variable functions: f = f i + f j + f z k. Theorem If f : D R n R, with n = 2, 3, is a differentiable function and u is a unit vector, then, D u f = ( f ) u.

The gradient vector and directional derivatives. Eample Find the gradient vector at an point in the domain of the function f (, ) = 2 + 2. Solution: The gradient is f = f, f, that is, f = 2, 2. Remark: f = 2r, with r =,. z f(,) = + 2 2 f D f u Directional derivatives and gradient vectors (Sect. 14.5). Directional derivative of functions of two variables. Partial derivatives and directional derivatives. Directional derivative of functions of three variables. The gradient vector and directional derivatives. Properties of the the gradient vector.

Properties of the the gradient vector. Remark: If θ is the angle between f and u, then holds D u f = f u = f cos(θ). The formula above implies: The function f increases the most rapidl when u is in the direction of f, that is, θ = 0. The maimum increase rate of f is f. The function f decreases the most rapidl when u is in the direction of f, that is, θ = π. The maimum decrease rate of f is f. The function f does not change along level curve or surfaces, that is, D u f = 0. Therefore, f is perpendicular to the level curves or level surfaces. Properties of the the gradient vector. Eample Find the direction of maimum increase of the function f (, ) = 2 /4 + 2 /9 at an arbitrar point (, ), and also at the points (1, 0) and (0, 1). Solution: The direction of maimum increase of f is the direction of its gradient vector: f = 2, 2 9 At the points (1, 0) and (0, 1) we obtain, respectivel, f =. 1 2, 0. f = 0, 2. 9

Properties of the the gradient vector. Eample Given the function f (, ) = 2 /4 + 2 /9, find the equation of a line tangent to a level curve f (, ) = 1 at the point P 0 = (1, 3 3/2). Solution: We first verif that P 0 belongs to the level curve f (, ) = 1. This is the case, since 1 4 + (9)(3) 4 The equation of the line we look for is 1 9 = 1. r(t) = 1, 3 3 + t v, v, 2 where v = v, v is tangent to the level curve f (, ) = 1 at P 0. Properties of the the gradient vector. Eample Given the function f (, ) = 2 /4 + 2 /9, find the equation of a line tangent to a level curve f (, ) = 1 at the point P 0 = (1, 3 3/2). Solution: Therefore, v f at P 0. Since, f = 2, 2 9 Therefore, ( f ) P0 = 1 2, 2 9 3 3 1 = 2 2, 1. 3 0 = v ( f ) P0 1 2 v = 1 3 v v = 2, 3. The line is r(t) = 1, 3 3 2 + t 2, 3.

Properties of the the gradient vector. f 3 3 1 2 f 1 2 f f = 1 2, 0, f = 0, 2, r(t) = 1, 3 3 + t 2, 3. 9 2 Further properties of the the gradient vector. Theorem If f, g are differentiable scalar valued vector functions, g 0, and k R an constant, then holds, 1. (kf ) = k ( f ); 2. (f ± g) = f ± g; 3. (fg) = ( f ) g + f ( g); ( f ) ( f ) g f ( g) 4. = g g 2.

Tangent planes and linear approimations (Sect. 14.6). Review: Differentiable functions of two variables. The tangent plane to the graph of a function. The linear approimation of a differentiable function. Bounds for the error of a linear approimation. The differential of a function. Review: Scalar functions of one variable. Scalar functions of more than one variable. Review: Differentiable functions of two variables. Definition Given a function f : D R 2 R and an interior point ( 0, 0 ) D, let L be the linear function L(, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ) + f ( 0, 0 ). The function f is called differentiable at ( 0, 0 ) iff the function f is approimated b the linear function L near ( 0, 0 ), that is, f (, ) = L(, ) + ɛ 1 ( 0 ) + ɛ 2 ( 0 ) where the functions ɛ 1 and ɛ 2 0 as (, ) ( 0, 0 ). Theorem If the partial derivatives f and f of a function f : D R 2 R are continuous in an open region R D, then f is differentiable in R.

Review: Differentiable functions of two variables. Eample Show that the function f (, ) = 2 + 2 is differentiable for all (, ) R 2. Furthermore, find the linear function L, mentioned in the definition of a differentiable function, at the point (1, 2). Solution: The partial derivatives of f are given b f (, ) = 2 and f (, ) = 2, which are continuous functions. Therefore, the function f is differentiable. The linear function L at (1, 2) is L(, ) = f (1, 2) ( 1) + f (1, 2) ( 2) + f (1, 2). That is, we need three numbers to find the linear function L: f (1, 2), f (1, 2), and f (1, 2). These numbers are: f (1, 2) = 2, f (1, 2) = 4, f (1, 2) = 5. Therefore, L(, ) = 2( 1) + 4( 2) + 5. Tangent planes and linear approimations (Sect. 14.6). Review: Differentiable functions of two variables. The tangent plane to the graph of a function. The linear approimation of a differentiable function. Bounds for the error of a linear approimation. The differential of a function. Review: Scalar functions of one variable. Scalar functions of more than one variable.

The tangent plane to the graph of a function. Remark: The function L(, ) = 2( 1) + 4( 2) + 5 is a plane in R 3. We usuall write down the equation of a plane using the notation z = L(, ), that is, z = 2( 1) + 4( 2) + 5, or equivalentl 2( 1) + 4( 2) (z 5) = 0. This is a plane passing through P 0 = (1, 2, 5) with normal vector n = 2, 4, 1. Analogousl, the function L(, ) = f ( 0, 0 )( 0 ) + f ( 0, 0 )( 0 ) + f ( 0, 0 ) is a plane in R 3. Using the notation z = L(, ) we obtain f ( 0, 0 )( 0 ) + f ( 0, 0 )( 0 ) (z f ( 0, 0 )) = 0. This is a plane passing through P 0 = ( 0, 0, f ( 0, 0 )) with normal vector n = f ( 0, 0 ), f ( 0, 0 ), 1. The tangent plane to the graph of a function. Theorem The plane tangent to the graph of a differentiable function f : D R 2 R at the point ( 0, 0 ) is given b Proof L(, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ) + f ( 0, 0 ). The plane contains the point P 0 = ( 0, 0, f ( 0, 0 )). We onl need to find its normal vector n. f (, ) = z n L(,) Plane that approimates P0 f (, ) at P 0 P 0 0

The tangent plane to the graph of a function. The vector n normal to the plane L(, ) is a vector perpendicular to the surface z = f (, ) at P 0 = ( 0, 0 ). f (, ) = z n L(,) Plane that approimates P0 f (, ) at P 0 P 0 This surface is the level surface F (,, z) = 0 of the function F (,, z) = f (, ) z. A vector normal to this level surface is its gradient F. That is, F = F, F, F z = f, f, 1. 0 Therefore, the normal to the tangent plane L(, ) at the point P 0 is n = f ( 0, 0 ), f ( 0, 0 ), 1. Recall that the plane contains the point P 0 = ( 0, 0, f ( 0, 0 )). The equation for the plane is f ( 0, 0 )( 0 ) + f ( 0, 0 )( 0 ) (z f ( 0, 0 )) = 0. The tangent plane to the graph of a function. Summar: We have shown that the linear L given b L(, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) is the plane tangent to the graph of f at ( 0, 0 ). f (, ) = z L(,) Plane that approimates f (, ) at (, ) 0 0 Plane that does not z L (, ) approimate f (,) near ( 0, 0 ). 1 f(,) (, ) 0 0 Function f is differentiable at (, ). 0 0 Function f is not differentiable at ( 0, 0 ). Remark: The graph of a differentiable function f : D R 2 R is approimated b the tangent plane L at ever point in D.

The tangent plane to the graph of a function. Eample Show that f (, ) = arctan( + 2) is differentiable and find the plane tangent to f (, ) at (1, 0). Solution: The partial derivatives of f are given b f (, ) = 1 1 + ( + 2) 2, f (, ) = 2 1 + ( + 2) 2. These functions are continuous in R 2, so f (, ) is differentiable at ever point in R 2. The plane L(, ) at (1, 0) is given b L(, ) = f (1, 0)( 1) + f (1, 0)( 0) + f (1, 0), where f (1, 0) = arctan(1) = π/4, f (1, 0) = 1/2, f (1, 0) = 1. Then, L(, ) = 1 2 ( 1) + + π 4. Tangent planes and linear approimations (Sect. 14.6). Review: Differentiable functions of two variables. The tangent plane to the graph of a function. The linear approimation of a differentiable function. Bounds for the error of a linear approimation. The differential of a function. Review: Scalar functions of one variable. Scalar functions of more than one variable.

The linear approimation of a differentiable function. Definition The linear approimation of a differentiable function f : D R 2 R at the point ( 0, 0 ) D is the plane L(, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ) + f ( 0, 0 ). Eample Find the linear approimation of f = 17 2 4 2 at (2, 1). Solution: L(, ) = f (2, 1)( 2) + f (2, 1)( 1) + f (2, 1). We need three numbers: f (2, 1), f (2, 1), and f (2, 1). These are: f (2, 1) = 3, f (2, 1) = 2/3, and f (2, 1) = 4/3. Then the plane is given b L(, ) = 2 3 ( 2) 4 ( 1) + 3. 3 Tangent planes and linear approimations (Sect. 14.6). Review: Differentiable functions of two variables. The tangent plane to the graph of a function. The linear approimation of a differentiable function. Bounds for the error of a linear approimation. The differential of a function. Review: Scalar functions of one variable. Scalar functions of more than one variable.

Bounds for the error of a linear approimation. Theorem Assume that the function f : D R 2 R has first and second partial derivatives continuous on an open set containing a rectangular region R D centered at the point ( 0, 0 ). If M R is the upper bound for f, f, and f in R, then the error E(, ) = f (, ) L(, ) satisfies the inequalit E(, ) 1 2 M ( 0 + 0 ) 2, where L(, ) is the linearization of f at ( 0, 0 ), that is, L(, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ) + f ( 0, 0 ). Bounds for the error of a linear approimation. Eample Find an upper bound for the error in the linear approimation of f (, ) = 2 + 2 at the point (1, 2) over the rectangle R = {(, ) R 2 : 1 < 0.1, 2 < 0.1} Solution: The second derivatives of f are f = 2, f = 2, f = 0. Therefore, we can take M = 2. Then the formula E(, ) 1 2 M ( 0 + 0 ) 2, implies E(, ) ( 1 + 2 ) 2 < (0.1 + 0.1) 2 = 0.04, that is E(, ) < 0.04. Since f (1, 2) = 5, the percentage relative error 100 E(, )/f (1, 2) is bounded b 0.8%

Tangent planes and linear approimations (Sect. 14.6). Review: Differentiable functions of two variables. The tangent plane to the graph of a function. The linear approimation of a differentiable function. Bounds for the error of a linear approimation. The differential of a function. Review: Scalar functions of one variable. Scalar functions of more than one variable. Review: Differential of functions of one variable. Definition The differential at 0 D of a differentiable function f : D R R is the linear function df () = L() f ( 0 ). Remark: The linear approimation of f () at 0 is the line given b L() = f ( 0 ) ( 0 ) + f ( 0 ). Therefore f() df () = f ( 0 ) ( 0 ). L() Denoting d = 0, f( ) 0 df f df = f ( 0 ) d. 0 d =

Differential of functions of more than one variable. Definition The differential at ( 0, 0 ) D of a differentiable function f : D R 2 R is the linear function df (, ) = L(, ) f ( 0, 0 ). Remark: The linear approimation of f (, ) at ( 0, 0 ) is the plane L(, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ) + f ( 0, 0 ). Therefore df (, ) = f ( 0, 0 ) ( 0 ) + f ( 0, 0 ) ( 0 ). Denoting d = 0 and d = ( 0 ) we obtain the usual epression df = f ( 0, 0 ) d + f ( 0, 0 ) d. Therefore, df and L are similar concepts: The linear approimation of a differentiable function f. Differential of functions of more than one variable. Eample Compute the df of the function f (, ) = ln(1 + 2 + 2 ) at the point (1, 1). Evaluate this df for d = 0.1, d = 0.2. Solution: The differential of f at ( 0, 0 ) is given b df = f ( 0, 0 )d + f ( 0, 0 )d. The partial derivatives f and f are given b f (, ) = 2 1 + 2 + 2, f (, ) = 2 1 + 2 + 2. Therefore, f (1, 1) = 2/3 = f (1, 1). Then df = 2 3 d + 2 3 d. Evaluating this differential at d = 0.1 and d = 0.2 we obtain df = 2 3 1 10 + 2 3 2 10 = 2 3 3 10 df = 1 5.

Differential of functions of more than one variable. Eample Use differentials to estimate the amount of aluminum needed to build a closed clindrical can with internal diameter of 8cm and height of 12cm if the aluminum is 0.04cm thick. Solution: The data of the problem is: h 0 = 12cm, r 0 = 4cm, dr = 0.04cm and dh = 0.08cm. The function to consider is the mass of the clinder, M = ρv, where ρ = 2.7gr/cm 3 is the aluminum densit and V is the volume of the clinder, V (r, h) = πr 2 h. The metal to build the can is given b M = ρ [ V (r + dr, h + dh) V (r, h) ], r = 4 ( recall dh = 2dr. ) 0 h = 12 0 dr = 0.04 Differential of functions of more than one variable. Eample Use differentials to estimate the amount of aluminum needed to build a closed clindrical can with internal diameter of 8cm and height of 12cm if the aluminum is 0.04cm thick. Solution: The metal to build the can is given b M = ρ [ V (r + dr, h + dh) V (r, h) ], A linear approimation to V = V (r + dr, h + dh) V (r, h) is dv = V r dr + V h dh, that is, dv = V r (r 0, h 0 )dr + V h (r 0, h 0 )dh. Since V (r, h) = πr 2 h, we obtain dv = 2πr 0 h 0 dr + πr 2 0 dh. Therefore, dv = 16.1 cm 3. Since dm = ρ dv, a linear estimate for the aluminum needed to build the can is dm = 43.47 gr.

Local and absolute etrema, saddle points (Sect. 14.7). Review: Local etrema for functions of one variable. Definition of local etrema. Characterization of local etrema. First derivative test. Second derivative test. Absolute etrema of a function in a domain. Review: Local etrema for functions of one variable. Recall: Main results on local etrema for f (): f() at f f f a ma. 0 < 0 b infl. 0 ± 0 c min. 0 > 0 d infl. = 0 ± 0 a b c d Remarks: Assume that f is twice continuousl differentiable. If 0 is local maimum or minimum of f, then f ( 0 ) = 0. If f ( 0 ) = 0 then 0 is a critical point of f, that is, 0 is a maimum or a minimum or an inflection point. The second derivative test determines whether a critical point is a maimum, minimum or an inflection point.

Local and absolute etrema, saddle points (Sect. 14.7). Review: Local etrema for functions of one variable. Definition of local etrema. Characterization of local etrema. First derivative test. Second derivative test. Absolute etrema of a function in a domain. Definition of local etrema for functions of two variables. Definition A function f : D R 2 R has a local maimum at the point (a, b) D iff holds that f (, ) f (a, b) for ever point (, ) in a neighborhood of (a, b). A function f : D R 2 R has a local minimum at the point (a, b) D iff holds that f (, ) f (a, b) for ever point (, ) in a neighborhood of (a, b). 01

Definition of local etrema for functions of two variables. Definition A differentiable function f : D R 2 R has a saddle point at an interior point (a, b) D iff in ever open disk in D centered at (a, b) there alwas eist points (, ) where f (, ) > f (a, b) and other points (, ) where f (, ) < f (a, b). Local and absolute etrema, saddle points (Sect. 14.7). Review: Local etrema for functions of one variable. Definition of local etrema. Characterization of local etrema. First derivative test. Second derivative test. Absolute etrema of a function in a domain.

Characterization of local etrema. First derivative test. Theorem If a differentiable function f has a local maimum or minimum at (a, b) then holds ( f ) (a,b) = 0, 0. Remark: The tangent plane at a local etremum is horizontal, since its normal vector is n = f, f, 1 = 0, 0, 1. Definition The interior point (a, b) D of a differentiable function f : D R 2 R is a critical point of f iff ( f ) (a,b) = 0, 0. Remark: Critical points include local maima, local minima, and saddle points. Characterization of local etrema. First derivative test. Eample Find the critical points of the function f (, ) = 2 2 Solution: The critical points are the points where f vanishes. Since f = 2, 2, the onl solution to f = 0, 0 is = 0, = 0. That is, (a, b) = (0, 0). Remark: Since f (, ) 0 for all (, ) R 2 and f (0, 0) = 0, then the point (0, 0) must be a local maimum of f. Eample Find the critical points of the function f (, ) = 2 2 Solution: Since f = 2, 2, the onl solution to f = 0, 0 is = 0, = 0. That is, we again obtain (a, b) = (0, 0).

Characterization of local etrema. Second derivative test. Theorem Let (a, b) be a critical point of f : D R 2 R, that is, ( f ) (a,b) = 0, 0. Assume that f has continuous second derivatives in an open disk in D with center in (a, b) and denote D = f (a, b) f (a, b) [ f (a, b) ] 2. Then, the following statements hold: If D > 0 and f (a, b) > 0, then f (a, b) is a local minimum. If D > 0 and f (a, b) < 0, then f (a, b) is a local maimum. If D < 0, then f (a, b) is a saddle point. If D = 0 the test is inconclusive. Notation: The number D is called the discriminant of f at (a, b). Characterization of local etrema. Second derivative test. Eample Find the local etrema of f (, ) = 2 2 and determine whether the are local maimum, minimum, or saddle points. Solution: We first find the critical points: f = 2, 2 ( f ) (a,b) = 0, 0 iff (a, b) = (0, 0). The onl critical point is (a, b) = (0, 0). We need to compute D = f (a, b) f (a, b) [ f (a, b) ] 2. Since f (0, 0) = 2, f (0, 0) = 2, and f (0, 0) = 0, we get D = ( 2)(2) = 4 < 0 saddle point at (0, 0).

Characterization of local etrema. Second derivative test. Eample Is the point (a, b) = (0, 0) a local etrema of f (, ) = 2 2? Solution: We first verif that (0, 0) is a critical point of f : f (, ) = 2 2, 2 2, ( f ) (0,0) = 0, 0, therefore, (0, 0) is a critical point. Remark: The whole aes = 0 and = 0 are critical points of f. We need to compute D = f (a, b) f (a, b) [ f (a, b) ] 2. Since f (, ) = 2 2, f (, ) = 2 2, and f (, ) = 4, we obtain f (0, 0) = 0, f (0, 0) = 0, and f (0, 0) = 0, hence D = 0 and the test is inconclusive. Characterization of local etrema. Second derivative test. Eample Is the point (a, b) = (0, 0) a local etrema of f (, ) = 2 2? Solution: From the graph of f = 2 2 is simple to see that (0, 0) is a local minimum: (also a global minimum.) z

Local and absolute etrema, saddle points (Sect. 14.7). Review: Local etrema for functions of one variable. Definition of local etrema. Characterization of local etrema. First derivative test. Second derivative test. Absolute etrema of a function in a domain. Absolute etrema of a function in a domain. Definition A function f : D R 2 R has an absolute maimum at the point (a, b) D iff f (, ) f (a, b) for all (, ) D. A function f : D R 2 R has an absolute minimum at the point (a, b) D iff f (, ) f (a, b) for all (, ) D. Remark: Local etrema need not be the absolute etrema. f() Absolute maimum Local maimum Local and absolute minimum a c d b f() No absolute etrema Remark: Absolute etrema ma not be defined on open intervals. Local minimum Local maimum a c d b

Review: Functions of one variable. Theorem Ever continuous functions f : [a, b] R R, with a < b R alwas has absolute etrema. f() Absolute maimum Local maimum Local and absolute minimum a c d b Recall: Intervals [a, b] are bounded and closed sets in R. The set [a, b] is closed, since the boundar points belong to the set, and it is bounded, since it does not etend to infinit. Recall: On open and closed sets in R n. Definition A set S R n, with n N, is called open iff ever point in S is an interior point. The set S is called closed iff S contains its boundar. A set S is called bounded iff S is contained in ball, otherwise S is called unbounded. 000000000 111111111 000000000000 111111111111 0000000000000 1111111111111 00000000000000 11111111111111 00000000000000 11111111111111 000000000000000 111111111111111 000000000000000 111111111111111 00000000000000 11111111111111 00000000000000 11111111111111 000000000000 111111111111 000000000 111111111 open and bounded 0000000 1111111 00000000 11111111 0000000000 1111111111 0000000000 1111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 000000000000 111111111111 00000000000 11111111111 00000000000 11111111111 00000000000 11111111111 0000000000 1111111111 0000000000 1111111111 00000000 11111111 000111 bounded closed and unbounded 000000000000000000000 111111111111111111111 0000000000000000000000000 1111111111111111111111111 000000000000000000000000000 111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 000000000000000000000000000000 111111111111111111111111111111 000000000000000000000000000000 111111111111111111111111111111 0000000000000000000000000000000 1111111111111111111111111111111 000000000000000000000000000000 111111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000000 11111111111111111111111111111 00000000000000000000000000 11111111111111111111111111 0000000000000000000000 1111111111111111111111 00000 11111 closed and bounded Theorem If f : D R 2 R is continuous in a closed and bounded set D, then f has an absolute maimum and an absolute minimum in D.

Absolute etrema on closed and bounded sets. Problem: Find the absolute etrema of a function f : D R 2 R in a closed and bounded set D. Solution: (1) Find ever critical point of f in the interior of D and evaluate f at these points. (2) Find the boundar points of D where f has local etrema, and evaluate f at these points. (3) Look at the list of values for f found in the previous two steps. If f ( 0, 0 ) is the biggest (smallest) value of f in the list above, then ( 0, 0 ) is the absolute maimum (minimum) of f in D. Absolute etrema on closed and bounded sets. Eample Find the absolute etrema of the function f (, ) = 3 + + 2 on the closed domain given in the Figure. Solution: (1) We find all critical points in the interior of the domain: f = ( 1), ( + 2) = 0, 0 ( 0, 0 ) = ( 2, 1). Since ( 2, 1) does not belong to the domain, we discard it. 4 II 1 I III 5 (2) Three segments form the boundar of D: Boundar I: The segment = 0, [1, 5]. We select the end points (1, 0), (5, 0), and we record: f (1, 0) = 2 and f (5, 0) = 2. We look for critical point on the interior of Boundar I: Since g() = f (, 0) = 3, so g = 1 0. No critical points in the interior of Boundar I.

Absolute etrema on closed and bounded sets. Eample Find the absolute etrema of the function f (, ) = 3 + + 2 on the closed domain given in the Figure. Solution: Boundar II: The segment = 1, [0, 4]. We select the end point (1, 4) and we record: f (1, 4) = 14. We look for critical point on the interior of Boundar II: Since g() = f (1, ) = 3 + 1 + 2 = 2 + 3, so g = 3 0. No critical points in the interior of Boundar II. 4 II 1 I III 5 Boundar III: The segment = + 5, [1, 5]. We look for critical point on the interior of Boundar III: Since g() = f (, + 5) = 3 + ( + 5) + 2( + 5). We obtain g() = 2 + 2 + 13, hence g () = 2 + 2 = 0 implies = 1. So, = 4, and we selected the point (1, 4), which was alread in our list. No critical points in the interior of Boundar III. Absolute etrema on closed and bounded sets. Eample Find the absolute etrema of the function f (, ) = 3 + + 2 on the closed domain given in the Figure. 4 II 1 I III 5 Solution: (3) Our list of values is: f (1, 0) = 2 f (1, 4) = 14 f (5, 0) = 2. We conclude: Absolute maimum at (1, 4), Absolute minimum at (5, 0).

A maimization problem with a constraint. Eample Find the maimum volume of a closed rectangular bo with a given surface area A 0. Solution: This problem can be solved b finding the local maimum of an appropriate function f. The function f is obtained as follows: Recall the functions volume and area of a rectangular bo with verte at (0, 0, 0) and sides, and z: V (,, z) = z, A(,, z) = 2 + 2z + 2z. Since A(,, z) = A 0, we obtain z = A 0 2, that is 2( + ) f (, ) = A 0 2 2 2. 2( + ) A maimization problem with a constraint. Eample Find the maimum volume of a closed rectangular bo with a given surface area A 0. Solution: We must find the critical points of f (, ) = A 0 2 2 2. 2( + ) f = 2A 0 2 4 2 2 8 3 4( + ) 2, f = 2A 0 2 4 2 2 8 3 4( + ) 2. The conditions f = 0 and f = 0 and 0, 0 impl } A 0 = 2 2 + 4, A 0 = 2 2 =. Recall z = A 0 2 + 4, 2( + ), so, z = A 0 2 2 4 =. Therefore, 0 = 0 = z 0 = A 0 /6.