w6-C - Math 20C Multivariable Calculus Lecture 16 1 ' $...

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Unformatted text preview: Math 20C Multivariable Calculus Lecture 16 1 ' $ Directional derivative and gradient vector Definition of directional derivative. (Sec. 14.6) Directional derivative and partial derivatives. Gradient vector. Geometrical meaning of the gradient. Slide 1 & % ' The directional derivative generalizes the partial derivatives to any direction Definition 1 The directional derivative of the function f (x, y) at the point (x0 , y0 ) in the direction of a unit vector u = ux , uy if $ Slide 2 1 [f (x0 + ux t, y0 + uy t) - f (x0 , y0 )] , t0 t if the limit exists. Du f (x0 , y0 ) = lim Particular cases: u = 1, 0 = i, then Di f (x0 , y0 ) = fx (x0 , y0 ). & u = 0, 1 = j, then Dj f (x0 , y0 ) = fy (x0 , y0 ). % Slide 4 Slide 3 Math 20C Multivariable Calculus & & ' ' The directional derivative generalizes the partial derivatives to any direction Recall the definition of partial derivatives x x z z f(x,y) f(x,y) (a,b) (a,b) u Df u Lecture 16 |u|=1 fx fy y y % $ % $ 2 Math 20C Multivariable Calculus Lecture 16 3 ' $ |u| = 1 implies that t is the distance between the points (x, y) = (x0 + ux t, y0 + uy t) and (x0 , y0 ) d = | x - x0 , y - y0 |, Slide 5 = |t| |u|, = |t|. = | ux t, uy t |, The directional derivative of f (x, y) at (x0 , y0 ) along u is the pointwise rate of change of f with respect to the distance along the line parallel to u passing through (x0 , y0 ). & % ' $ Here is a useful formula to compute directional derivative Theorem 1 If f (x, y) is differentiable and u = ux , uy is a unit vector, then Du f (x0 , y0 ) = fx (x0 , y0 ) ux + fy (x0 , y0 ) uy . The proof is based in the chain rule, case 1 Slide 6 & % Math 20C Multivariable Calculus Lecture 16 4 ' $ Proof of the theorem Chain rule case 1, for x(t) = x0 + ux t, y(t) = y0 + uy t. Then, z(t) = f (x(t), y(t)). Slide 7 On the one hand, dz dt 1 = lim [z(t) - z(0)], t0 t 1 = lim [f (x0 + ux t, y0 + uy t) - f (x0 , y0 )], t0 t = Du f (x0 , y0 ). % t=0 & ' Proof of the theorem (Cont.) On the other hand, dz dx dy (t) = fx (x(t), y(t)) (t) + fy (x(t), y(y)) (t), dt dt dt = fx (x(t), y(t))ux + fy (x(t), y(t))uy , $ Slide 8 then, dz dt Therefore, Du f (x0 , y0 ) = fx (x0 , y0 )ux + fy (x0 , y0 )uy . & % = fx (x0 , y0 )ux + fy (x0 , y0 )uy . t=0 Math 20C Multivariable Calculus Lecture 16 5 ' Example about how to compute a directional derivative Let f (x, y) = sin(x + 2y). Compute the directional derivative of f (x, y) at (4, -2) in the direction = /6. $ Slide 9 Also u = cos(), sin() , fx = cos(x + 2y), then Du f (x, y) Du f (4, -2) = = u= 3/2, 1/2 . fy = 2 cos(x + 2y), cos(x + 2y)ux + 2 cos(x + 2y)uy , 3 + 1. 2 & % ' $ Directional derivatives can be defined on functions of 2, 3 or more variables Definition 2 (functions of 3 variables) The directional derivative of the function f (x, y, z) at the point (x0 , y0 , z0 ) in the direction of a unit vector u = ux , uy , uz is Du f (x0 , y0 , z0 ) = lim t0 Slide 10 1 [f (x0 + ux t, y0 + uy t, z0 + uz t) - f (x0 , y0 , z0 )] , t if the limit exists. & % Math 20C Multivariable Calculus Lecture 16 6 ' $ The same useful theorem we had in 2 variable functions Slide 11 Theorem 2 If f (x, y, z) is differentiable and u = ux , uy , uz is a unit vector, then Du f (x0 , y0 , z0 ) = fx (x0 , y0 , z0 ) ux + fy (x0 , y0 , z0 ) uy + fz (x0 , y0 , z0 )uz . & % ' $ The directional derivative can be written in terms of a dot product In the case of 2 variable functions: Slide 12 with Du f = fx ux + fy uy = ( f ) u, f = f x , fy . In the case of 3 variable functions: Du f = fx ux + fy uy + fz uz = ( f ) u, with & f = f x , fy , fz . % Math 20C Multivariable Calculus Lecture 16 7 ' $ We introduce the gradient vector for functions of 2 or 3 variables Definition 3 Let f (x, y, z) be a differentiable function. Then, f (x, y, z) = fx (x, y, z), fy (x, y, z), fz (x, y, z) , is called the gradient of f (x, y, z). In 2 variables: Alternative notation: Slide 13 f (x, y) = fx (x, y), fy (x, y) . f = fx i + fy j + fz k. & % ' The useful theorem now has the following form $ Theorem 3 Let f (x, y, z) be differentiable function. Then, Du f (x) = ( f (x)) u. Slide 14 with |u| = 1. z f(x,y) = x 2 + y 2 y f x y x & Duf % Math 20C Multivariable Calculus Lecture 16 8 ' $ Gradient vector The gradient vector has two main properties: Slide 15 It points in the direction of the maximum increase of f , and | f | is the value of the maximum increase rate. f is normal to the level surfaces. & % ' $ Here is the first property of the gradient vector Theorem 4 Let f be a differentiable function of 2 or 3 variables. Fix P0 D(f ), and let u be an arbitrary unit vector. Then, the maximum value of Du f (P0 ) among all possible directions is | f (P0 )|, and it is achieved for u parallel to f (P0 ). Slide 16 & % Math 20C Multivariable Calculus Lecture 16 9 ' $ The proof of the first property Du f (P0 ) = = = ( f (P0 )) u, | f (P0 )| |u| cos(), | f (P0 )| cos(). Slide 17 But -1 cos() 1 implies -| f (P0 )| Du f (P0 ) | f (P0 )|. And Du f (P0 ) = | f (P0 )|, = 0 u is parallel f (P0 ). & % ' $ Here is the second property of the gradient vector, in the case of 3 variable functions Slide 18 Theorem 5 Let f (x, y, z) be a differentiable at P0 . Then, f (P0 ) is orthogonal to the plane tangent to a level surface containing P0 . & % Math 20C Multivariable Calculus Lecture 17 10 ' $ Proof of the second property Let r(t) be any differentiable curve in the level surface f (x, y, z) = k. - - Assume that r(t = 0) = OP 0 . Then, 0 = = = df , dt dy dz dx fx + fy + fz , dt dt dt dxr f (r(t)) (t). dt Slide 19 But (dr)/(dt) is tangent to the level surface for any choice of r(t). Therefore r f (r(t = 0)) (t = 0) = 0 dt implies that f (P0 ) is orthogonal to the level surface. & % ' $ Local and absolute maxima, minima, and inflection points Slide 20 Definitions of local extrema. (Sec. 14.7) Characterization of local extrema. Absolute extrema on closed and bounded sets. Typical exercises. & % Math 20C Multivariable Calculus Lecture 17 11 ' $ Recall the main results on local extrema for f (x) f(x) at Slide 21 a b c a b c d x f max. infl. min. infl. f 0 0 f <0 >0 = 0 0 = 0 0 d & % ' The main cases of local extrema for f (x, y) $ Slide 22 & !! !! % Math 20C Multivariable Calculus Lecture 17 12 ' $ The intuitive notions of local extrema can be written precisely as follows Definition 4 (Local maximum) A function f (x, y) has a local maximum at (a, b) D(f ) f (x, y) f (a, b) for all (x, y) near (a, b). Definition 5 (Local minimum) A function f (x, y) has a local minimum at (a, b) D(f ) f (x, y) f (a, b) for all (x, y) near (a, b). & % Slide 23 ' $ The tangent plane to the graph of f at a local max-min is horizontal Theorem 6 Let f (x, y) be differentiable at (a, b). If f has a local maximum or minimum at (a, b) then f (a, b) = 0, 0 . Recall: n = fx , fy , -1 = 0, 0, -1 . The converse is not true: It could be a saddle point & % Slide 24 Math 20C Multivariable Calculus Lecture 17 13 ' $ Stationary points include local maxima, minima, and saddle points Definition 6 (Stationary point) Let f (x, y) be a differentiable function at (a, b). If f (a, b) = 0, 0 , then the point (a, b) is called a stationary point of f . Stationary point are located where the gradient vector vanishes & % Slide 25 ' Theorem 7 (Second derivative test) Let (a, b) be a stationary point of f (x, y), that is, f (a, b) = 0. Assume that f (x, y) has continuous second derivatives in a disk with center in (a, b). Introduce the quantity D = fxx (a, b)fyy (a, b) - [fxy (a, b)]2 . $ Slide 26 If D > 0 and fxx (a, b) > 0, then f (a, b) is a local minimum. If D > 0 and fxx (a, b) < 0, then f (a, b) is a local maximum. If D < 0, then f (a, b) is a saddle point. & If D = 0 the test is inconclusive. % Math 20C Multivariable Calculus Lecture 17 14 ' $ Find the local extrema of f (x, y) = y 2 - x2 f = -2x, 2y , Slide 27 fxx (0, 0) = -2, fyy (0, 0) = 2, fxy (0, 0) = 0, f = 0, 0 at (0, 0). D = (-2)(2) = -4 < 0 & saddle point at (0, 0). % ' $ Is (0,0) a local extrema of f (x, y) = y 2 x2 ? f (x, y) = 2xy 2 , 2yx2 , Slide 28 f (0, 0) = 0, 0 at (0, 0). fxx (x, y) = 2y 2 , fxx (0, 0) = 0, fyy (x, y) = 2x2 , fyy (0, 0) = 0, fxy (x, y) = 4xy, fxy (0, 0) = 0, So D = 0 and the test is inconclusive. & % Math 20C Multivariable Calculus Lecture 17 15 ' From the graph of f = x2 y 2 is easy to see that (0, 0) is a global minimum z $ Slide 29 y x & % ' $ Find the maximum volume of a closed rectangular box with a given surface area A0 V (x, y, z) = xyz, Slide 30 A(x, y, z) = 2xy + 2xz + 2yz. But A(x, y, z) = A0 , then z= Find A0 - 2xy , 2(x + y) V (x, y) = A0 xy - 2x2 y 2 . 2(x + y) V (x0 , y0 ) = 0, 0 . A0 /6. % The result is x0 = y0 = z0 = & Math 20C Multivariable Calculus Lecture 17 16 ' $ Local extrema need not be the absolute extrema f(x) f(x) Slide 31 a c b x a c b x Absolute extrema may not be defined on open intervals & % ' Continuous functions f (x) on intervals [a, b] always have absolute extrema f(x) $ Slide 32 a c b x Intervals [a, b] are bounded and closed sets in I R Because they do not extend to infinity, and the boundary points belong to the set. & % $% $% $% $% $% $% $% $% $% $% $% $% $% $% $% $% $% $%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% $%$% $%$%$% % % % % % % % % % % % % % % % % % $$$$$$$$$$$$$$$$$ %$%%$%%$%%$%%$%%$%%$%%$%%$%%$%%$%%$%%$%%$%%$%%$% $$$$$$$$$$$$$$$$ $%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% $%$% $$%%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% $%$% %$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$ % % % % % % % % % % % % % % % % $$$$$$$$$$$$$$$$$% $%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% $%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% $%$% % % % % % % % % % % % % % % % % $$$%% %%%%%%%%%%%%%%%%%% $$$$$$$$$$$$$$$$$ $%$$%$$%$$%$$%$$%$$%$$%$$%$$%$$%$$%$$%$$%$$%$$%$ $%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% $%$% $$$$$$$$$$$$$$$$$ $$%%$% % % % % % % % % % % % % % % % % % $%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$%$% y # # # # # # # # # # # # # # # # # ################ ""#"""""""""""""""" ""## "#""#""#""#""#""#""#""#""#""#""#""#""#""#""#""#" "#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"# "# "#"#"# ""#""#""#""#""#""#""#""#""#""#""#""#""#""#""#""# # # # # # # # # # # # # # # # # #"################# """"""""""""""""" "#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"# "#"# "#"#"# # # # # # # # # # # # # # # # # # """"""""""""""""" #"##"##"##"##"##"##"##"##"##"##"##"##"##"##"##"# """""""""""""""" "#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"# "#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"# "# "#"#"# ""#""#""#""#""#""#""#""#""#""#""#""#""#""#""#""# # # # # # # # # # # # # # # # # #"################# """"""""""""""""""# "#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"#"# "#"# y Slide 34 Slide 33 Math 20C Multivariable Calculus ' & & ' in P always contains both points in D and points not in D. 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Slide 37 ( f (x, y) = 0. No second derivative test needed.) Find the extrema (max. and min.) values of f on the boundary of D. The biggest (smallest) of the previous steps is the absolute maximum (minimum). & % ' $ Here is a typical exercise Find the absolute extrema of f (x, y) = 4x + 6y - x2 - y 2 , on D = {(x, y) I 2 , 0 x 4, 0 y 5} R Absolute minimum: (4, 0), (0, 0). Absolute maximum: (2, 3). Slide 38 & % Math 20C Multivariable Calculus Lecture 18 20 ' $ Lagrange's multipliers Example of the method. Maximization of functions subject to constraints. Examples. Generalization to more than one constraint. Slide 39 & % ' Example: Find the rectangle of biggest area with fixed perimeter P0 One way to solve the problem is: A(x, y) = xy, P0 = P (x, y) = 2x + 2y, $ then y = P0 /2 - x, and replace it in A(x, y), Slide 40 A(x) = P0 x - x2 . 2 The stationary points of this function are 0 = A (x) = P0 P0 P0 - 2x, x = , y= . 2 4 4 P0 . 4 So the answer is the square of side x=y= & % Math 20C Multivariable Calculus Lecture 18 21 ' $ Idea behind the Lagrange multipliers method y y = A/x y = - x + P/2 y Slide 41 x x Level curves of A = xy, & Level curves of the constraint P = 2x + 2y. % ' The gradient vectors of A(x, y) and of the constraint P = 2x + 2y are parallel at the solution y y = A/x $ Slide 42 y = -x + P/2 & x % Math 20C Multivariable Calculus Lecture 18 22 ' The same problem solved with the Lagrange multipliers method $ Find the maximum of A(x, y) = xy subject to the constraint P (x, y) = 2x + 2y = P0 . One has to find the (x, y) such that Slide 43 A(x, y) = P (x, y), P (x, y) = P0 , with = 0. From the first equation one has y, x = 2, 2 , x = 2, y = 2. Then the constraint P0 = 2x + 2y implies that P0 = 8, so the answer is x=y= P0 . 4 & % ' $ Lagrange multipliers method can be summarized as follows: The extrema values of f (x, y) subject to the constraint g(x, y) = k can be obtained as follows: Slide 44 Find all solutions (x0 , y0 ) and of the equations f (x0 , y0 ) = g(x0 , y0 ), g(x0 , y0 ) = k. Evaluate f at every solution (x0 , y0 ). The largest and smallest values are respectively the maximum and minimum values of f subject to the constraint g = k. & % Math 20C Multivariable Calculus Lecture 18 23 ' $ Lagrange multipliers method for functions of three variables The extrema values of f (x, y, z) subject to the constraint g(x, y, z) = k can be obtained as follows: Slide 45 Find all solutions (x0 , y0 , z0 ) and of the equations f (x0 , y0 , z0 ) = g(x0 , y0 , z0 ), g(x0 , y0 , z0 ) = k. Evaluate f at every solution (x0 , y0 , z0 ). The largest and smallest values are respectively the maximum and minimum values of f subject to the constraint g = k. & % ' Example: Find the rectangular box of maximum volume for fixed area. $ The function is V (x, y, z) = xyz. The constraint function is A(x, y, z) = 2xy + 2xz + 2yz. The constraint is A(x, y, z) = A0 . Find the (x, y, z) solutions of V = A, Slide 46 These equations are: A = A0 . yz = 2(z + y), xz = 2(x + z), xy = 2(x + y), A0 = 2(xy + xz + zy). p & The solution is x = y = z = A0 /6. % Math 20C Multivariable Calculus Lecture 18 24 ' Example: Find the extrema values of f (x, y) = x2 + y 2 /4 in the circle x2 + y 2 = 1 Then, f (x, y) = x2 + y 2 /4, and g(x, y) = x2 + y 2 . The equations are: f = g, g = 1, 2x, y/2 = 2x, 2y , x2 + y 2 = 1. $ Slide 47 Which imply x = x, y/2 = 2y, x + y = 1. The solutions are: P = (0, 1), and P = (1, 0). Then: f (0, 1) = 1/4, absolute minimum in the circle. 2 2 (1 - )x = 0, (1/4 - )y = 0, & f (1, 0) = 1, absolute maximum in the circle. % ' Generalization to two constraints $ The extrema values of f (x, y, z) subject to the constraints g(x, y, z) = k1 and h(x, y, z) = k2 can be obtained as follows: Find all solutions (x0 , y0 , z0 ) and of the equations Slide 48 f (x0 , y0 , z0 ) = g(x0 , y0 , z0 ) + h(x0 , y0 , z0 ), g(x0 , y0 , z0 ) = k1 , h(x0 , y0 , z0 ) = k2 . Evaluate f at every solution (x0 , y0 , z0 ). The largest and smallest values are respectively the maximum and minimum values of f subject to the constraint g = k1 and h = k2 . & % ...
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