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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 )] , t→0 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)], t→0 t 1 = lim [f (x0 + ux t, y0 + uy t) − f (x0 , y0 )], t→0 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 θ = π/6. $ Slide 9 Also u = cos(θ), sin(θ) , 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 t→0 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(θ), | f (P0 )| cos(θ). Slide 17 But −1 ≤ cos(θ) ≤ 1 implies −| f (P0 )| ≤ Du f (P0 ) ≤ | f (P0 )|. And Du f (P0 ) = | f (P0 )|, ⇔ θ = 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 & % ' 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. Definition 7 (Bounded and closed sets in I 2 ) R $ Slide 32 • A set D ⊂ I 2 is bounded if it can be contained in a R disk. • A point P ∈ I 2 is a boundary point of a set D if R every disk with center in P always contains both points in D and points not in D . • A set D ∈ I 2 is closed if it contains all its boundary R points. & % Math 20C Multivariable Calculus Lecture 17 17 ' $ Here are examples of different types of sets y y closed and Slide 33 open and bounded bounded x unbounded x closed and bounded & % ' $ Continuous functions on bounded and closed sets always have absolute extrema Slide 34 Theorem 8 If f (x, y ) is continuous in a closed and bounded set D ⊂ I 2 , then f has an absolute maximum R and an absolute minimum in D . & % Math 20C Multivariable Calculus Lecture 18 18 ' $ Suggestions to find absolute extrema of f (x, y ) in a closed and bounded set • Find every stationary point of f . Slide 35 ( 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 36 & % Math 20C Multivariable Calculus Lecture 18 19 ' $ Lagrange multipliers • Example of the method. Slide 37 • Lagrange multipliers method: Maximization of functions subject to constraints. • Examples. • Generalization to more than one constraint. & % ' Example Find the rectangle of biggest area with fixed perimeter P0 . $ The usual way to solve the problem is: Slide 38 A(x, y ) = xy, P0 = P (x, y ) = 2x + 2y, then y = P0 /2 − x, and replace it in A(x, y ), P0 x − x2 . 2 The stationary points of this function are A(x) = & 0 = A (x) = P0 P0 P0 − 2x, ⇒ x = ,⇒ y= . 2 4 4 % Math 20C Multivariable Calculus Lecture 18 20 ' $ So the answer is the square of side Slide 39 x=y= P0 . 4 & % ' 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 40 A(x, y ) = λ P (x, y ), P (x, y ) = P0 , with λ = 0. From the first equation one has y , x = λ 2, 2 , ⇒ x = 2λ, y = 2λ. & Then the constraint P0 = 2x + 2y implies that P0 = 8λ, so the answer is P0 . x=y= 4 % Math 20C Multivariable Calculus Lecture 18 21 ' $ Lagrange multipliers method Theorem 9 The extrema values of f (x, y ) subject to the constraint g (x, y ) = k can be obtained as follows: Slide 41 • Find all solutions (x0 , y0 ) and λ of the equations f (x0 , y0 ) = λ 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 . & % ' $ Lagrange multipliers method Theorem 10 The extrema values of f (x, y, z ) subject to the constraint g (x, y, z ) = k can be obtained as follows: Slide 42 • Find all solutions (x0 , y0 , z0 ) and λ of the equations f (x0 , y0 , z0 ) = λ 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 . & % Math 20C Multivariable Calculus Lecture 18 22 ' Example of Lagrange multipliers method • 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 . Slide 43 Find the (x, y, z ) solutions of V A These equations are: yz xz xy = = = = 2λ(z + y ), 2λ(x + z ), 2λ(x + y ), A0 . = = λ A, A0 . & 2(xy + xz + zy ) % ' $ Slide 44 The solution is x = y = z = p A0 /6. & % Math 20C Multivariable Calculus Lecture 18 23 ' Example of Lagrange multipliers method • Find the extrema values of f (x, y ) = x2 + y 2 /4 in the circle x2 + y 2 = 1. Then, f (x, y, z ) = x2 + y 2 /4, and g (x, y ) = x2 + y 2 . The equations are: f = = λ g, 1, ⇒ ⇒ λx, 2λy, 1. 2x, y/2 x +y 2 2 $ = = λ 2x, 2y , 1. Slide 45 g Which imply x y/2 x +y 2 2 = = = ⇒ ⇒ (1/4 − λ)y (1 − λ)x = = 0, 0, The solutions are: P = (0, ±1), and P = (±1, 0). Then: f (0, ±1) = 1/4, absolute minimum in the circle. & f (±1, 0) = 1, absolute maximum in the circle. % ' Generalization to two constraints $ Theorem 11 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 λ of the equations Slide 46 f (x0 , y0 , z0 ) = λ g (x0 , y0 , z0 ) + µ 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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This note was uploaded on 12/26/2009 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.

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