= w. w u. u ; u + w. x x. z z. y y. v + w. . Remark. The formula stated above is very important in the theory of. surface integral.

Transcription

1 1 Chain rules 2 Directional derivative 3 Gradient Vector Field 4 Most Rapid Increase 5 Implicit Function Theorem, Implicit Differentiation 6 Lagrange Multiplier 7 Second Derivative Test

2 Theorem Suppose that w = f (x, y, z) is a differentiable function, where x = x(u, v), y = y(u, v), z = z(u, v), where the coordinate functions are parameterized by differentiable functions Then the composite function w(u, v) = f ( x(u, v), (u, v), (u, v) ) is a differentiable function in u and v, such that the partial functions are given by w u w v = w x x u + w y y u + w z z u ; = w x x v + w y y v + w z z v Remark The formula stated above is very important in the theory of surface integral

3 Theorem (Chain Rule for Coordinate Changes) Suppose that s = f (x, y, z) is a differentiable function, where x = x(u, v, w), y = y(u, v, w), z = z(u, v, w), where the coordinate functions are parameterized by differentiable functions in variables u, v and w Then the composite function S(u, v, w) = f ( x(u, v, w), (u, v, w), (u, v, w) ) is a differentiable function in u, v and w, such that the partial functions are given by S u S v = S x x u + S y y u + S z z u ; = S x x v + S y y v + S z z v ; S w = S x x w + S y y w + S z z w Remark The formula stated above is very important in the theory of inverse function theory and integration theory

4 Example In spherical coordinates, we have the parameters (ρ, θ, ϕ) to represent (x, y, z) as follows: x = ρ sin ϕ cos θ, y = ρ sin ϕ sin θ, z = ρ cos ϕ, with ρ 0, 0 θ 2π, and 0 ϕ π Define S(x, y, z) = x 2 + y 2 + z 2 Evaluate the partial derivative S ρ in two ways Solution (i) Since S(ρ sin ϕ cos θ, ρ sin ϕ sin θ, ρ cos ϕ) = ρ, so S ρ = 1 for any choices of parameters involved (ii) The second method is to apply chain rule S x = S y = and S z = x = sin ϕ cos θ, x x 2 +y 2 +z 2 ρ = ρ (ρ sin ϕ cos θ) = sin ϕ cos θ, y = sin ϕ sin θ, y x 2 +y 2 +z 2 ρ = ρ (ρ sin ϕ sin θ) = sin ϕ sin θ, z = cos ϕ, z x 2 +y 2 +z 2 ρ = ρ (ρ cos ϕ) = cos ϕ And S ρ = S x x ρ + S y y ρ + S z z ρ = (sin ϕ cos θ) 2 + (sin ϕ sin θ) 2 + (cos ϕ) 2 = (sin 2 ϕ)(cos 2 θ + sin 2 θ ) + cos 2 ϕ = 1

5 Theorem (Chain Rule of 2-variables) Suppose that f (x, y)nd is a real valued function defined on the planar domain D, and that r(t) = x(t)i + y(t)j is a curve in the domain D Then we obtain a real-valued function g(t) = f (x(t), y(t)) which is a function of t Then the derivative of g is given by g (t) = d f ( f (x(t), y(t)) ) = dt x f x (r(t))x (t) + f y (r(t))y (t) dx (r(t)) dt + f dy (r(t)) y dt =

6 Theorem Chain Rule of 3-variables Suppose that f (x, y, z) is real valued function defined on the domain D which is part of R 3, and that x = x(t), y = y(t) and z = z(t) is a curve in the domain D One can think of the a particle moving in domain D, and its position is given by (x(t), y(t), z(t)) changing with respect to t, so it traces out a path in domain D given by r(t) = x(t)i + y(t)j + z(t)k Then we obtain a real-valued function g(t) = f (x(t), y(t), z(t)) Then the chain rule tells us that g (t) = d ( f (x(t), y(t), z(t)) ) dt = f dx x (r(t)) dt + f dy y (r(t)) dz (r(t)) dt = f x (r(t))x (t) + f y (r(t))y (t) + f z (r(t))z (t) dt + f z (chain rule)

7 Partial Derivatives Suppose that z = f (x, y) is a function defined in a domain D, and P(a, b) is a point in D Recall that the partial derivatives f (a + h, b) f (a, b) f x (a, b) = lim, and h 0 h f (a, b + k) f (a, b) f y (a, b) = lim k 0 k The limits are taken along the coordinate axes

8 Directional Derivative Through the point P(a, b) we choose any direction u = (h, k) = hi + kj, then we consider the straight line through the point P along the direction u given by r(t) = (a + ht, b + kt), and the rate of change of g(t) = f (r(t)) at t = 0 is z g f ( r(t) ) f ( r(0) ) f (a+th,b+tk) f (a,b) (0) = lim t 0 t = lim t 0 t T P(x,y,z ) y x

9 Directional Derivative Through the point P(a, b) we choose any direction u = (h, k) = hi + kj, then we consider the straight line through the point P along the direction u given by r(t) = (a + ht, b + kt), and the rate of change of g(t) = f (r(t)) at t = 0 is g f ( r(t) ) f ( r(0) ) f (a+th,b+tk) f (a,b) (0) = lim t 0 t = lim t 0 t Suppose that f is differentiable, then it follows from the (multivariate) chain rule that g (0) = f dx x dt + f dy y dt = f f (P)h + x y (P)k = ( f x (a, b)i + f y (a, b)j ) (hi + kj) = f (a, b) u, where f is the vector-valued function f x i + f y j, called the gradient of f at the point (x, y) Note g (0) only depends of the choice of the curve through P(a, b) with tangent direction r (0) only

10 In order to simplify the notation more, one requires the directional vector u to be an unit vector Definition (Directional derivative) The resulting derivative g (0) is called the directional derivative D u f of f along the direction u, and hence we write D u f = f u to represent the rate of the change of f in the unit direction u Remark In general, if f = f (x 1,, x n ) is a function of n variables, one define (i) the gradient of f to be f = ( f f x, 1, x n ), and (ii) the directional derivative D u f by D u f = f u

11 Example Evaluate the directional derivative f (x, y) = xe y + cos(xy) at the point P 0 (2, 0) in the direction 3i 4j Remark The blue curve is the level curve of f at different values Solution Let u = 3i 4i = i 4 5j And f (2, 0) = (f x (2, 0), f y (2, 0)) = (e y y sin(xy), xe y x sin(xy)) (x,y)=(2,0) = (1, 2) It follows that D u f (2, 0) = f (2, 0) u = = 1

12 Proposition The greatest rate of change of a scalar function f, ie, the maximum directional derivative, takes place in the direction of, and has the magnitude of, the vector f Proof For any direction v, the directional derivative of f along the direction v at a point P in the domain of f, is given by v D v (P) := f (P), = f cos θ, where θ is the angle between v the vectors f (P) and v Hence D v (P) attains maximum (minimum) value if and only if cos θ = 1 ( 1), if and only if f (P) ( f (P) ) is parallel to v In this case, we have D v (P) = f ( f )

13 Example (a) Find the directional derivative of f (x, y, z) = 2x 3 y 3y 2 z at P(1, 2, 1) in a direction v = 2i 3j + 6k (b) In what direction from P is the directional derivative a maximum? (c) What is the magnitude of the maximum directional derivative? Solution (a) f (P) = 6x 2 yi + (2x 3 6yz)j 3y 2 k (1,2, 1) = 12i + 14j 12k at P Then the directional derivative of f along the direction v is given by D v f = f v v = 12i + 14j 12k, 2i 3j+6k = 90 7 (b) D v f (P) is maximum(minimum) v ( v) is parallel to f (P) = 12i + 14j 12k (c) The maximum magnitude of D v f (P) is given by f (P) f = f (P) = 12i + 14j 12k = f (P) f (P) = 22 = f (P) 2

14 Proposition Let C : r(t) = x(t)i + y(t)j + z(t)k be a curve lying on the level surface S : f (x, y, z) = c for some c, ie c = f (r(t)) = f ( x(t), y(t), z(t) ) for all t Then the gradient vector f of f is always perpendicular to the tangent vector r (t) at r(t) for all t ie f (r(t)) r (t) Proof Define the composite function g(t) = f ( x(t), y(t), z(t) ), it follows from the given condition that g(t) = f ( x(t), y(t), z(t) ) = c is a constant function, so one can differentiate the identity c = g(t) = f ( x(t), y(t), z(t) ), so 0 = g (t) = f x dx dt + f y dy dt + f z dx dt = f (r(t)), r (t) for all t So f r (t) at r(t) for all t

15 Proposition Let f (x, y, z) be a differentiable function defined in R 3, and S : f (x, y, z) = c be a level surface for some constant c, ie S = { (x, y, z) f (x, y, z) = c } Suppose that P(a, b, c) be a point on S such that the gradient vector f (a, b, c) of f at point P(a, b, c) is not zero, then the equation of the tangent plane of S at P is given by < f (a, b, c), (x a, y b, z c) >= 0, ie f x (a, b, c)(x a) + f y (a, b, c)(y b) + f z (a, b, c)(z c) = 0 Remark For any given level surface S defined by a scalar function f, the tangent plane of S at any P of S is spanned by the tangent vector of the curve contained in S The result above tells us that the normal direction to the tangent plane of S at any point P of S is parallel to f (P)

16 Let f (x, y) be a differentiable function defined on xy-plane For any real number k, recall the level level of f for k is given by the set { (x, y) f (x, y) = k } When the value k changes, the level curve changes gradually Let f (x, y) = x 2 7xy + 2y 2 defined on xy-plane The blue curves represent the level curves C k : f (x, y) = k of various values c And the red arrows represent the gradient vector field f (a, b) = ( f x (a, b), f y (a, b) ) which is normal to the tangent vector to level curve C a at P(a, b) of various values k Proposition Let C k : f (x, y) = k be a fixed level curve with a point P(a, b) in C k If f (a, b) = (0, 0), then the equation of the tangent line of C k at P is given by f (a, b) (x a, y b) = 0, ie f x (a, b)(x a) + f y (a, b)(y b) = 0

17 Example Let F(x, y, z) = x α + y α + z α, where α is non-zero number Determine the equation of the tangent plane of the level surface S : F(x, y, z) = k of some point P(a, b, c) in S, where k is a positive constant Solution The normal direction N of the tangent plane of S at P is parallel to F(x, y, z) = α(x α 1, y α 1, z α 1 ) evaluated at P(a, b, c) So N = (a α 1, b α 1, c α 1 ), it follows that the equation of the tangent plane of S at P(a, b, c) is given by 0 = N, (x a, y b, z c) = a α 1 (x a) + b α 1 (y b) + c α 1 (z c), So the equation of the tangent plane of S at P is given by a α 1 x + b α 1 y + c α 1 z = a α + b α + c α = F(a, b, c) = k

18 Example Show that the surface S : x 2 2yz + y 3 = 4 is perpendicular to any member of the family of surfaces S a : x = (2 4a)y 2 + az 2 at the point of intersection P(1, 1, 2) Solution Let the defining equations of level surfaces S, S a be F(x, y, z) = x 2 2yz + y 3 4 = 0 and G(x, y, z) = x (2 4a)y 2 az 2 = 0 Then F(x, y, z) = 2xi + (3y2 2z)j 2yk, and G(x, y, z) = 2xi 2(2 4a)yj 2azk Thus, the normals to the two surfaces at P(1, 1, 2) are given by N 1 = 2i j + 2k, N 2 = 2i + 2(2 4a)j 4ak Since N 1 N 2 = (2)(2) 2(2 4a) (2)(4a) = 0, it follows that N 1 and N 2 are perpendicular for all a, and so the required result follows

19 Implicit Functions Given a relation between two variables expressed by an equation of the form f (x, y) = k, we often want to solve for y That is, for each given x in some interval, we expect to find one and only one value y = φ(x) that satisfies the relation The function φ is thus implicit in the relation; geometrically, the locus of the equation f (x, y) = k is a level curve in the (x, y)-plane that serves as the graph of the function y = φ(x) Example Let C : x 2 + y 2 = 1 be a level curve defined by a function f (x, y) = x 2 + y 2 Is it possible that this level curve C in xy-plane is given by the graph of some "nice" function? If y = g(x), then we have 1 = x 2 + (g(x)) 2, and hence g(x) 2 = 1 x 2, so g(x) = ± 1 x 2 Though we find out a possible representation of y = g(x), which is usually called "explicit function," in fact g not differentiable at x = ±1 On the contrary, we call y is defined implicitly by f (x, y) = c

20 The most familiar example of an implicitly defined function is provided by the equation f (x, y) = x 2 + y 2 The locus or level curve f (x, y) = k is a circle of radius k if k > 0, we can view it as the graph of two different functions, y = φ ± (x) = ± k x 2 we can take either P + (0, + k) or P ( 0, k) as a fixed point of the level curve, so that φ ± defines a function respectively such that (i) the graph passes through the point P ± (0, ± k), and (ii) the graph of f lies completely on the level curve, ie all the points (x, φ ± (x)) lies on the level curve, f (x, φ ± (x)) = k for all x dom(f )

21 The explicit functions φ ± defined by means of implicit function f (x, y) = k, satisfy (i) the graph passes through the point P ± (0, ± k), and (ii) the graph of f lies completely on the level curve, ie all the points (x, φ ± (x)) lies on the level curve, f (x, φ ± (x)) = k for all x dom(f ) Thing completely fails if we chose the point P( k, 0), the reason is that a function can takes on only one value, though we can write down y = + k x 2 for k x k, but the graph can not be extended to any bigger domain to meet the second condition (ii) Moreover, the function y = + k x 2 does not have any derivative at x = k, which checked directly

22 Implicit Function Theorem I Let C : f (x, y) = k be a level curve defined by a differentiable scalar function f of 2 variables Suppose P(a, b) is a point in the domain of f such that f y (P) = 0, then there exists δ > 0 and a differentiable function g defined in an interval I = (a δ, a + δ) such that (i) f (x, g(x)) = c for all x I with g(a) = b; ie y = g(x) is an explicit function defined by the level curve C; and (ii) g (x) = f x(x, g(x)) for all x I f y (x, g(x)) Remark (i) In general, we can t write down the explicit function g (ii) one can interchange the role of x and y, if f x (P) = 0

23 Remark Recall that the level surface associated to a scalar function f and a fixed number k, is the set { (x, y, z) f (x, y, z) = k } In general, this set is not expected to have any nice condition However, we have the following important Implicit Function Theorem II Let S : F(x, y, z) = k be a level surface defined by a differentiable scalar function F, and suppose that P(a, b, c) is a point on the level surface, ie F(a, b, c) = k Suppose that F z (P) = 0, then there exists δ > 0 and a differentiable function z = g(x, y) defined on the open disc B( (a, b), δ) such that (i) F(x, y, g(x, y)) = k for all (x, y) B( (a, b), δ), with g(a, b) = c; and (ii) g x (x, y) = F x(x, y, g(x, y) ) g F z (x, y, g(x, y) ) y (x, y) = F y(x, y, g(x, y) ) for all F z (x, y, g(x, y) ) (x, y) B( (a, b), δ)

24 Implicit Function Theorem II Let S : F(x, y, z) = k be a level surface defined by a differentiable scalar function F, and suppose that P(a, b, c) is a point on the level surface, ie F(a, b, c) = k Suppose that F z (P) = 0, then there exists δ > 0 and a differentiable function z = g(x, y) defined on the open disc B( (a, b), δ) such that (i) F(x, y, g(x, y)) = k for all (x, y) B( (a, b), δ), with g(a, b) = c; and (ii) g x (x, y) = F x(x, y, g(x, y) ) g F z (x, y, g(x, y) ) y (x, y) = F y(x, y, g(x, y) ) for all F z (x, y, g(x, y) ) (x, y) B( (a, b), δ) Remark Differentiate F(x, y, g(x, y)) = k with respect to x and y respectively by means of chain rule, we have F F g (x, y, g(x, y)) + (x, y, g(x, y)) (x, y) = (k) = 0, and the x z x x result follows

25 Let z = z(x, y) be implicitly defined by ze xz = 2z + y + 1 Find z x at the point (x, y, z) = (0, 0, 1) Solution Write z(x, y) instead of z, and then differentiate the identity z(x, y)e xz(x,y) = 2z(x, y) + y + 1 with respect to x, we have z x e xz + ze xz (xz x + z) = (ze xz ) x = (2z + y + 1) x = 2z x, hence z x (e xz + xze xz 2) = z 2 e xz At (x, y, z) = (0, 0, 1), we have z x ( ) = ( 1) 2, ie z x (0, 0) = 1

26 Example Suppose that the implicit function given by the level surface S : F(x, y, z) = 0 defines the following explicit functions: x = x(y, z), y = y(x, z) and z = x(x, y), where F is a differentiable function Then x y y z z x = Solution It follows from the implicit function theorem that x y = F y F x, for all (x, y, z) in S Similarly, we have y z = F z, and z F y x = F x, for F z all (x, y, z) in S It follows that x y y z z ( x = F ) ( y F ) ( z F ) x = 1, for all (x, y, z) in S F x F y F z

27 Theorem Let r(t) = (x(t), y(t), z(t)) be a curve on the level surface S : f (x, y, z) = c, prove that the tangent vector r (t) of the curve r(t) is normal to the gradient f at the point of S Consequently, f is the normal vector of the tangent plane of level surface S at any point P(x, y, z) of S Proof The result follows easily from differentiate the identity c = f ( x(t), y(t), z(t) ) for all the t in the domain of r with the help of chain rule, so 0 = d dt (c) = d ( f (x(t), y(t), z(t)) ) = dt f dx x dt + f dy y dt + f dz dx = f ( z dt dt, dy dt, dz dt ) = f r (t), so f is normal to the tangent vector r (t) of the curve

28 Example Determine the extremum of the function z = z(x, y) defined implicitly by the equation 3x 2 + 2y 2 + z 2 + 8yz z + 8 = 0 Solution Define F(x, y, z) = 3x 2 + 2y 2 + z 2 + 8yz z + 8, so the function z = z(x, y) is in fact the graph of the level surface S associated to the equation F(x, y, z) = 0, or sometimes we just denote it by S : F(x, y, z) = 0 It follows that F(x, y, z(x, y)) = 0, for all (x, y) in the (unspecified) domain of z(x, y), in fact we just think of the equality as an identity in x and y So differentiate with respect to x and y respectively by means of chain rule, we have 0 = x (0) = F x ( F(x, y, z(x, y)) ) = x x x + F z z x = F z x + F z x, so one has z x (x, y) = F x(x, y, z(x, y)) F z (x, y, z(x, y)) = F x and F z z y (x, y) = F y(x, y, z(x, y)) F z (x, y, z(x, y)) = F y One should notice that the F z assumption F z = 0 for all (x, y) in the domain of z = z(x, y) is necessary, which one can obtain explicitly if F z is known

29 Example Determine the extremum of the function z = z(x, y) defined implicitly by the equation 3x 2 + 2y 2 + z 2 + 8yz z + 8 = 0 Solution Let F(x, y, z) = 3x 2 + 2y 2 + z 2 + 8yz z + 8 so z x = F x 6x z =, and F z 2z + 8y 1 y = F y 4y + 8z = F z 2z + 8y 1 To locate the extremum value of z = z(x, y), one need its two partial derivatives z x and z y vanish, ie (6x, 4y + 8z) = (0, 0) where (x, y, z) is a point of the level surface S : F(x, y, z) = 0 Hence, x = 0, and y = 2z Then 0 = F(0, 2z, z) = 2( 2z) 2 + z 2 + 8( 2z)z z + 8 = 7z 2 z + 8 = (7z + 8)(z 1) so z = 1 or 7 8 Hence P(0, 2, 1) or Q(0, 16 7, 8 7 ) are the only critical point of the function z = z(x, y), however, z = z(x, y) is not explicitly determined yet One can determine use the quadratic formula to express z in terms of x and y, and then one can see that z max = 1 and z min = 8 7 Remark In the last part, we skip some details, but the gap can be filled in after we learn the second derivative test

30 Theorem (Lagrange multiplier) Let f (x, y) and g(x, y) be functions with continuous first-order partial derivatives If the maximum (minimum) value of f subject to the condition (constraint) given by a level curve C : g(x, y) = 0 occurs at a point P where f (P) = 0, then f (P) = λ g(p) for some real number λ Remarks 1 The last condition just means that these two vectors f (P) and g(p) are parallel, in other words, at the point where f attains maximum, the level curve of f will tangent to the constraint curve 2 The last equation f (P) = λ g(p) gives a necessary condition for finding the point P, though λ is also an unknown: f x (x, y) = λg x (x, y), f y (x, y) = λg y (x, y), g(x, y) = 0 3 The similar condition f (P) = λ g(p) works for functions of any variables, and the constant λ is called a multiplier

31 Example Determine min value of x 2 + y 2 subject to the constraint xy = 1 Solution Let f (x, y) = x 2 + y 2 be the objective function, and g(x, y) = xy be the constraint with the level curve given C : g(x, y) = 1 Though C is not a bounded set, one can put more restriction x 2 + y 2 M with the result curve C M which is closed and bounded As C M is closed and bounded in R 2, then the continuous function f attains its minimum on C M at some point in C M In fact, the minimum value always occurs exactly at the same two points One apply the Lagrange multiplier to locate the minimum that f (x, y) = λ g(x, y) at those extremum points

32 Example Determine min value of x 2 + y 2 subject to the constraint xy = 1 Solution (One should know that it is only a necessary condition, but not sufficient one) Hence we have: (2x, 2y) = (λy, λx), and xy = 1 From the last equation, one knows that x = 0 and y = 0, so 2x = λy, and then λ = 2x/y Substituting, we have 2y = (2x/y)x and hence y 2 = x 2, ie y = ±x But xy = 1, so x = y = ±1 and the possible points for the extreme values of f are (1, 1) and ( 1, 1) The minimum value is f (1, 1) = f ( 1, 1) = 2 Remark Here there is no maximum value for f, since the constraint xy = 1 allows x or y to become arbitrarily large, and hence f (x, y) = x 2 + y 2 can be made arbitrarily large

33 Steps of implementing Lagrange multipliers To find the maximum and minimum values of f (x, y, z) subject to the constraint defined by the level surface S : g(x, y, z) = k Suppose that these extreme values exist and on the surface S, which is related to the condition of S 1 Find all values of x, y, z and λ such that f x (x, y, z) = λg x (x, y, z) (1) f = λ g f y (x, y, z) = λg y (x, y, z) (2) f z (x, y, z) = λg z (x, y, z) (3) and g(x, y, z) = k (4) 2 Evaluate f at all the points (x, y, z) that result from step (a) The largest of these values is the maximum value of f ; the smallest is the minimum value of f

34 Example Use Lagrange multipliers to find the point (x, y, z) at which x 2 + y 2 + z 2 is minimal subject to x + 2y + 3z = 1 Solution Let f (x, y, z) = x 2 + y 2 + z 2, and g(x, y, z) = x + 2y + 3z be the objective function and the constraint function respectively We want to locate the point P(x, y) on the plane x + 2y + 3z = 1, such that f = λ g for some λ, ie (2x, 2y, 2z) = λ(1, 2, 3), and so 1 = x + 2y + 3z = λ 2 + 2λ + 3 3λ 2 = 7λ, ie λ = 7 1 And hence (x, y, z) = ( λ 2, λ, 3λ 2 ) = (1/14, 1/7, 3/14) Remark Why does the point (x, y, z) = (1/14, 1/7, 3/14) give the minimum of f? One can consider the moving point (x, y, z) = (3t + 1, 0, t) lying on the plane x + 2y + 3z = 1, then f (2t + 1, 0, t) = (2t + 1) ( t) 2 = (2t + 1) 2 + t 2 t 2 which does not have any maximum value However, one can prove that the absolute minimum value of f does exist by means of Cauchy s inequality, and we skip the proof of this fact

35 Example A rectangular box is placed on the xy-plane so that one vertex at the origin, and the opposite vertex lies in the plane Ax + By + Cz = 1, where A, B and C are positive Find the maximum volume of such a box Solution It follows from the given condition that the box has dimension x y z, with x, y, z > 0 and satisfy Ax + By + Cz = 1 Then the volume V(x, y, z) = xyz, subject to the constraint D = { (x, y, z) Ax + By + Cz = 1, and x, y, z 0 }, which is a closed and bounded subset of R 3, hence the volume function V attains both maximum and minimum The minimum volume is obviously 0; and we use Lagrange multiplier to find the maximum volume as follows(yz, xz, xy) = V(x, y, z) = λ (Ax + By + Cz 1) = (λa, λb, λc) If λ = 0, then one of x, y, and z will be zero, in this case, V(x, y, z) = 0, which is not maximum Assume λ = 0 so x 2 = xy xz yz AC = λc λb λa = λ BC A, ie x = BC A λ Similarly, we have B λ, and z = y = ( BC A A + B AC B AB + C AB C λ At last, we have 1 = Ax + By + Cz = ) λ C = 3 ABC λ, so λ = 1 9ABC Then

36 Example Let r(t) = (a + ht, b + kt) be the line in xy-plane passing through the point P(a, b) Let f be a function defined in a domain D containing P with continuous second order partial derivatives, and that P is a critical point of f ie f (P) = 0 Let g(t) = f (r(t)), (i) evaluate the second derivatives of g at t = 0; and (ii) the sign of g (0) provided that f xx (a, b) > 0 and f xx (a, b)f yy (a, b) (f xy (a, b)) 2 > 0 for (h, k) = (0, 0) Solution (i) Let A = f xx (a, b), C = f yy (a, b), B = (f xy (a, b) It follows from chain rule that g (t) = f x (a + ht, b + kt)h + f y (a + ht, b + kt)k, and hence g (t) = f xx (a + ht, b + kt)h 2 + f xy (a + ht, b + kt)hk + f yx (a + ht, b + kt)kh + f yy (a + ht, b + kt)k 2 In particular, at t = 0, g (0) = f xx (a, b)h 2 + 2f xy (a, b)hk + f yy (a, b)k 2 = Ah 2 + 2Bhk + Ck 2 (ii) As A = f xx (a, b) > 0, and AC B 2 > 0, then for s R, then l(s) = As 2 + 2Bs + C = 1 A (A2 s 2 + 2ABs + B 2 ) + C B 2 /A = 1 A (As + B)2 + AC B2 A AC B2 A > 0 So, and hence g (0) = Ah 2 + 2Bhk + Ck 2 = k 2 (A(h/k) 2 + 2Bh/k + C) = k 2 l( h k ) > 0 for all (h, k) R 2 with k = 0 If k = 0, then g (0) = Ah 2 > 0 for all h = 0

37 Proposition (Maximum-Minimum Test for Quadratic Functions) Let g(x, y) = Ax 2 + 2Bxy + Cy 2, where A, B, C are constants 1 If AC B 2 > 0, and A > 0, [respectively A < 0], then g(x, y) has a minimum [respectively maximum] at (0, 0) 2 If AC B 2 < 0, then g(x, y) takes both positive and negative values near (0, 0), so (0, 0) is not a local extremum for g Proof To prove these assertions, we consider the two cases separately (1) ( If AC B 2 > 0, then A ( cannot be zero (why?), so g(x, y) = A x 2 + 2B A xy + A C y2) = A x 2 + 2B B2 A xy + y 2 C A 2 A y2 B2 y 2) A ( ) 2 2 = A x + A B y + 1 A (AC B 2 )y 2 Both terms above have the same sign as A, and they are both zero only when x + A B y = 0 and y = 0, ie (x, y) = (0, 0) Thus (0, 0) is a minimum point for g if A > 0 (since g(x, y) > 0 if (x, y) = (0, 0)) and a maximum point if A < 0 (since g(x, y) < 0 if (x, y) = (0, 0) ) This completes the proof of (1)

38 Proposition (Maximum-Minimum Test for Quadratic Functions) Let g(x, y) = Ax 2 + 2Bxy + Cy 2, where A, B, C are constants 1 If AC B 2 > 0, and A > 0, [respectively A < 0], then g(x, y) has a minimum [respectively maximum] at (0, 0) 2 If AC B 2 < 0, then g(x, y) takes both positive and negative values near (0, 0), so (0, 0) is not a local extremum for g Proof (2) If AC B 2 < 0 and A = 0, then formula (1) still applies, but now the terms on the right-hand side have opposite signs By suitable choices of x and y (try it!), we can make either term zero and the other nonzero If A = 0, then g(x, y) = y(2bx + Cy), so we can again achieve both signs Remark In case (2), (0, 0) is called a saddle point for g(x, y)

39 Theorem (Second Derivatives Test) Suppose the second partial derivatives of f (x, y) are continuous on a disk with center (a, b), and suppose that f (a, b) = (0, 0) ie (a, b) is a critical point of f Let D = D(a, b) = f xx (a, b)f yy (a, b) [f xy (a, b)] 2 1 If D > 0 and f xx (a, b) > 0, then f (a, b) is a local minimum; 2 If D > 0 and f xx (a, b) < 0, then f (a, b) is a local maximum; 3 If D < 0, then f (a, b) is neither a local maximum nor a local minimum

40 Example Determine the nature of the critical points of f (x, y) = x 3 + y 3 6xy Solution As f (x, y) = (3x 2 6y, 3y 2 6x), so x 2 = 2y and y 2 = 2x It follows that x 4 = 4y 2 = 4 2x = 8x, ie 0 = x 4 8x = x(x ) = x(x 2)(x 2 + 2x + 4) As (x 2 + 2x + 4) = (x + 1) > 0, we have x = 0 or x = 2 So 2y = 0 2 or 2 2, so the critical points of f are (0, 0) and (2, 2) Next we need to apply the 2nd derivative testand f xx = 6x, f yy = 6y, f xy = 6, and the discriminant (x, y) = f xx f yy f 2 xy = (6x)(6y) ( 6) 2 = 36(xy 1) And (0, 0) = 36 < 0, (2, 2) = 36(4 1) = 108 Hence (0, 0) is a saddle point of f, where (2, 2) is a local minimum point of f

41 Taylor s Formula for f (x, y) at the Point (a, b) Theorem Suppose f (x, y) and its partial derivatives through order n + 1 are continuous throughout an open rectangular region R centered at a point (a, b) Then, throughout R, f (a + h, b + k) = f (a, b) + (hf x + kf y ) (a,b) }{{} + Linear or 1st order approximation f (a + h, b + k) = f (a, b) + (hf x + kf y ) (a,b) + 1 2! (h2 f xx + 2hkf xy + k 2 f yy ) (a,b) + }{{} 2nd order approximation f (a + h, b + k) = f (a, b) + (hf x + kf y ) (a,b) + 2! 1 (h2 f xx + 2hkf xy + k 2 f yy ) (a,b) + 3! 1 (h3 f xxx + 3h 2 kf xxy + 2hk 2 f xyy + k 3 f yyy ) (a,b) + ( ) n + n! 1 h x + k y f (a,b) + 1 (n+1)! for some c (0, 1) ( h x + k y ) n+1 f (a+ch,b+ck)

42 Taylor s Theorem Suppose f (x, y) and its partial derivatives through order n + 1 are continuous throughout an open rectangular region R centered at a point (a, b) Then, throughout R, = f (a, b) + (hf x + kf y ) (a,b) + 2! 1 (h2 f xx + 2hkf xy + k 2 f yy ) (a,b) + 3! 1 (h3 f xxx + 3h 2 kf xxy + 2hk 2 f xyy + k 3 f yyy ) (a,b) + ( ) n ( ) n+1 + n! 1 h x + k y f (a,b) + 1 h (n+1)! x + k y f (a+ch,b+ck) for some c (0, 1) Remarks 1 The proof just applies the chain rule and the trick of n-th Taylor polynomial to the function g(t) = f (a + ht, b + kt) one variable 2 If one have an estimate the last term (in blue), for example an upper bound, then we can estimate the given function by means of polynomials in 2 variables 3 The theorem can be easily generalized to function of n variables for n 1 Though this topics is not treated in this book, but its application is important in other courses, so we put the result in this notes for the sake of students