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1 Tutorial Class V 3-10/10/ First Order Partial Derivatives;

2 Tutorial Class V 3-10/10/ First Order Partial Derivatives; 2 Application of Gradient;

3 Tutorial Class V 3-10/10/ First Order Partial Derivatives; 2 Application of Gradient; 3 Directional Derivative;

4 Tutorial Class V 3-10/10/ First Order Partial Derivatives; 2 Application of Gradient; 3 Directional Derivative; 4 Critical Points, and Gradient of Scalar Function;

5 Tutorial Class V 3-10/10/ First Order Partial Derivatives; 2 Application of Gradient; 3 Directional Derivative; 4 Critical Points, and Gradient of Scalar Function; 5 Maximum and Minimum;

6 Tutorial Class V 3-10/10/ First Order Partial Derivatives; 2 Application of Gradient; 3 Directional Derivative; 4 Critical Points, and Gradient of Scalar Function; 5 Maximum and Minimum; 6 Implicit Function and Its Partial Derivatives;

7 Tutorial Class V 3-10/10/ First Order Partial Derivatives; 2 Application of Gradient; 3 Directional Derivative; 4 Critical Points, and Gradient of Scalar Function; 5 Maximum and Minimum; 6 Implicit Function and Its Partial Derivatives; 7 Weekly Quiz

8 Let w = f (x, y), where x = x(r, θ) = r cos θ and y = y(r, θ) = r sin θ Consider the composite function w = f (r cos θ, r sin θ) as a function of variables r and θ Find w w and in terms of derivatives with r θ respect to θ and r Solution We first calculate some simple partial derivatives x r = x (r cos θ) = cos θ, r θ = (r cos θ) = r sin θ, and θ y r = y (r sin θ) = sin θ, r θ = (r sin θ) = r cos θ, and θ By means of chain rule, we have w r = w x x r + w y y r = f x (r cos θ, r sin θ) cos θ + f y (r cos θ, r sin θ) sin θ, and w θ = w x x θ + w y y θ = f x (r cos θ, r sin θ) r sin θ + f y (r cos θ, r sin θ) ( r cos θ) Remark The reason why we substitute (r cos θ, r sin θ) for (x, y) in the partial derivatives of f with respect to x and y is that we are thinking w θ and w r as functions of r and θ

9 Example Let w = f (x, y), where x = x(r, θ) = r cos θ and y = y(r, θ) = r sin θ Consider the composite function w = f (r cos θ, r sin θ) as a function of variables r and θ Express w x, w w y and ( x )2 + ( w y )2 in terms of w w r and θ Solution It follows from previous problem that 1 r w θ = sin θ f x(r cos θ, r sin θ) cos θ f y (r cos θ, r sin θ), and w r = cos θ f x (r cos θ, r sin θ) + sin θ f y (r cos θ, r sin θ) Now think of a system of linear equations in unknown f x and f y : { wθ r = sin θ f x cos θ f y, (1) w r = cos θ f x + sin θ f y (2) Eliminate f y from (1) and (2) with sin θ (1) + cos θ (2), so sin θ wθ r + cos θ w r = sin 2 θ f x + cos 2 θ f x = f x Eliminate f x from (1) and (2) with cos θ (1) + sin θ (2), so cos θ wθ r + sin θ w r = cos 2 θ f y + sin 2 θ f y = f y, and ) 2 ( + cos θ w r + cos θ w θ ) 2 r + sin θ w ( ) r = (sin 2 θ + cos 2 w θ) 2 θ + w 2 r 2 r = w2 θ + w 2 r 2 r f 2 x + f 2 y = ( sin θ wθ r

10 1 Find the gradient f of the function f (x, y, z) = ( 1, 0, 1) Solution f x (x, y, z) = x (3 x2 y 2 z 2 ) 1/2 ) = 1 2 (3 x2 y 2 z 2 ) 1/2 1 = x(3 x 2 y 2 z 2 ) 3/2 x = 3 x 2 y 2 z 2 ) 2 3 x 2 y 2 z 1 So f x ( 1, 0, 1) = = 1 ( 3 ( 1) 2 (0) 2 (1) 2 ) 3 Similarly, we have f y ( 1, 0, 1) = 0, and f z ( 1, 0, 1) = 1 f ( 1, 0, 1) = ( f x ( 1, 0, 1), f y ( 1, 0, 1), f z ( 1, 0, 1) ) = ( 1, 0, 1) 2 for at x (3 x2 y 2 z 2 )

11 Let f (x, y, z) = h(g(x, y, z)) where h is scalar function of 1 variable Then f = A h (x, y, z) g(x, y, z) B g (x, y, z) h(x, y, z) C g(x, y, z) h(x, y, z) D h (g(x, y, z)) g(x, y, z) E None of the above Solution Answer is D f (x, y, z) = h(g(x, y, z)) x x = h (g(x, y, z)) g(x, y, z) x = h (g(x, y, z))g x (x, y, z) Similarly y f (x, y, z) = h (g(x, y, z))g y (x, y, z) and z f (x, y, z) = h (g(x, y, z))g z (x, y, z) In this case, we have f (x, y, z) = h (g(x, y, z))(f x (x, y, z), f y (x, y, z), f z (x, y, z)) = h (g(x, y, z)) f (x, y, z)

12 Notation For any given f, one write f (x, y) = f x (x, y)i + f y (x, y)j = (f x, f y ) at point (x, y)

13 Notation For any given f, one write f (x, y) = f x (x, y)i + f y (x, y)j = (f x, f y ) at point (x, y) If f has 1st order partial derivatives on its domain, then (a, b) is a critical point of f

14 Notation For any given f, one write f (x, y) = f x (x, y)i + f y (x, y)j = (f x, f y ) at point (x, y) If f has 1st order partial derivatives on its domain, then (a, b) is a critical point of f f (a, b) = (0, 0) So one can think of this condition as a set of equations to determine the possible interior critical points

15 Determine the minimum of the function f (x, y) = xy, where R = { (x, y) x 2 + y 2 + xy 1 } Hint: 4x 2 + 4y 2 + 4xy = (x + 2y) 2 + 3x 2 For the boundary of the region R, let x 3 = 2 cos t, and x + 2y = sin t for 0 t 2π

16 Determine the minimum of the function f (x, y) = xy, where R = { (x, y) x 2 + y 2 + xy 1 } Hint: 4x 2 + 4y 2 + 4xy = (x + 2y) 2 + 3x 2 For the boundary of the region R, let x 3 = 2 cos t, and x + 2y = sin t for 0 t 2π Solution Interior critical points: f (x, y) = (0, 0) if and only if (y, x) = (0, 0), ie (x, y) = (0, 0), which lies in the region R

17 Determine the minimum of the function f (x, y) = xy, where R = { (x, y) x 2 + y 2 + xy 1 } Hint: 4x 2 + 4y 2 + 4xy = (x + 2y) 2 + 3x 2 For the boundary of the region R, let x 3 = 2 cos t, and x + 2y = sin t for 0 t 2π Solution Interior critical points: f (x, y) = (0, 0) if and only if (y, x) = (0, 0), ie (x, y) = (0, 0), which lies in the region R Boundary points: For any point point (x, y) on the boundary of R, we set the curve C : x(t) = 2 cos t, and y(t) = sin t cos t for 0 t 2π, 3 3 one has x 2 + y 2 + xy = 1 4 (3x2 + x 2 + 4xy + 4y 2 ) = 1 4 (x + 2y) x2 = 1 4 (4 sin2 t) cos2 t 3 = sin 2 t + cos 2 t = 1 So parameterized curve r(t) = (x(t), y(t)) describes all the boundary points of R Let ( sin t cos t 3 ) = 1 g(t) = x(t)y(t) = 2 cos t ( 1 ) 2 sin(2t π 3 4 ) (sin 2t 1+cos 2t 3 ) =

18 Determine the minimum of the function f (x, y) = xy, where R = { (x, y) x 2 + y 2 + xy 1 } Hint: 4x 2 + 4y 2 + 4xy = (x + 2y) 2 + 3x 2 For the boundary of the region R, let x 3 = 2 cos t, and x + 2y = sin t for 0 t 2π Solution Interior critical points: f (x, y) = (0, 0) if and only if (y, x) = (0, 0), ie (x, y) = (0, 0), which lies in the region R Boundary points: For any point point (x, y) on the boundary of R, we set the curve C : x(t) = 2 cos t, and y(t) = sin t cos t for 0 t 2π, 3 3 one has x 2 + y 2 + xy = 1 4 (3x2 + x 2 + 4xy + 4y 2 ) = 1 4 (x + 2y) x2 = 1 4 (4 sin2 t) cos2 t 3 = sin 2 t + cos 2 t = 1 So parameterized curve r(t) = (x(t), y(t)) describes all the boundary points of R Let ( sin t cos t 3 ) = 1 1+cos 2t ) g(t) = x(t)y(t) = 2 cos t (sin 2t = ( 1 ) 2 sin(2t π ) In fact, it follows from AM-GM ineq that f (x, y) = xy = 1 3 (2xy + xy) 1 3 (x2 + y 2 + xy) = 1 3, with equality holds if 2xy = x 2 + y 2 ie x = y Then it follows that 1 = x 2 + xy + y 2 = x 2 + x 2 + x 2 = 3x 1, so x 2 = 1 3 ie x = ± 1 3

19 Determine the minimum value of xyz subject to the condition 1 x + 1 y + 1 z = 1 with positive x, y, z Solution The condition is g(x, y, z) = xy + yz + zx xyz = 0 and let f (x, y, z) = xyz The extremum value(s) could be given by Lagrange multiplier f = λ g subject to the condition xy + yz + zx xyz = 0 So η(yz, xz, xy) = (y + z yz, x + z xz, x + y xy), where η = 1/λ Hence (η + 1)yz = y + z, (η + 1)xz = x + z, (η + 1)xy = x + y, so (η + 1) = 1 y + 1 z = 1 x + 1 z = 1 x + 1 y And then 2 = 2( 1 x + 1 y + 1 z ) = (η + 1) + (η + 1) + (η + 1) = 3(η + 1), ie η = 1 3 It follows from that 1 x = 1 y = 1 z = 1 2 (η + 1) = 1 3 Hence we have x = y = z = 1+η 2 = 3 Then the minimum value is f (3, 3, 3) = 27 Moreover, one can check that 1 g(x, x, ) = x + 1 x = 2 x + (1 2 x ) = 1, and x 1 2 x 1 f (x, x, ) = 1 2 x 2 x3 = (x 2)2 + 6(x 2) x 2 8 (x 2)2 for all x x > 2 Hence there is no maximum value of f subject to the constraint g(x, y, z) = 0

20 Example Let a, p, n, m be given positive numbers, find the global maximum of the function h(x, y, z) = x m y n z p subject to the constraint x + y + z = a and x, y, z > 0 Justify your reason Solution D = { (x, y, z) x + y + z = a and x, y, z 0 } is a triangular region, which is closed and bounded subset of R 3 The continuous polynomial h(x, y, z) will attain maximum at some point D However, h can not attain maximum on the boundary of D as at least one of x, y and z vanishes

21 Example Let a, p, n, m be given positive numbers, find the global maximum of the function h(x, y, z) = x m y n z p subject to the constraint x + y + z = a and x, y, z > 0 Justify your reason Solution D = { (x, y, z) x + y + z = a and x, y, z 0 } is a triangular region, which is closed and bounded subset of R 3 The continuous polynomial h(x, y, z) will attain maximum at some point D However, h can not attain maximum on the boundary of D as at least one of x, y and z vanishes First consider the new function f (x, y, z) = ln(x m y n z p ) = m ln x + n ln y + p ln z As logarithm function ln is an increasing function, f (x, y, z) attains its maximum if and only if h(x, y, z) does Let g(x, y, z) = x + y + z, then the constraint is given by the level surface g(x, y, z) = x + y + z = a By using the Lagrange s multiplier ( m x, n y, p z ) = (m ln x + n ln y + p ln z) = f (x, y, z) =

22 Example Let a, p, n, m be given positive numbers, find the global maximum of the function h(x, y, z) = x m y n z p subject to the constraint x + y + z = a and x, y, z > 0 Justify your reason Solution D = { (x, y, z) x + y + z = a and x, y, z 0 } is a triangular region, which is closed and bounded subset of R 3 The continuous polynomial h(x, y, z) will attain maximum at some point D However, h can not attain maximum on the boundary of D as at least one of x, y and z vanishes First consider the new function f (x, y, z) = ln(x m y n z p ) = m ln x + n ln y + p ln z As logarithm function ln is an increasing function, f (x, y, z) attains its maximum if and only if h(x, y, z) does Let g(x, y, z) = x + y + z, then the constraint is given by the level surface g(x, y, z) = x + y + z = a By using the Lagrange s multiplier ( m x, n y, p z ) = (m ln x + n ln y + p ln z) = f (x, y, z) = λ g(x, y, z) = λ (x + y + z) = λ(1, 1, 1) for some λ R, and hence (x, y, z) = ( m λ, λ n, p λ ) It follows from x + y + z = a that a = x + y + z = m λ + λ n + p λ = m+n+p maximum occurs at (x, y, z) = ( λ am m+n+p, an m+n+p, max value = a m+n+p m m n n p p /(m + n + p) m+n+p, and we know that ) and h has ap m+n+p

23 Example Let u(x, y, z) = ln(x 3 + y 3 + z 3 3xyz), prove that u x + u y + u z = x+y+z 3 on the domain where all these functions are well-defined Solution u x = = u y = 1 x 3 + y 3 + z 3 3xyz x (x3 + y 3 + z 3 3xyz) 3x 2 3yz x 3 + y 3 + z 3 3xyz, 3y 2 3xz x 3 + y 3 + z 3 3xyz, u z = u x + u y + u z = 3(x2 + y 2 + z 2 xy yz xz) x 3 + y 3 + z 3 3xyz 3z 2 3xy x 3 + y 3 + z 3 3xyz = 3 x + y + z The last equality follows from the algebraic identity: x 3 + y 3 + z 3 3xyz = (x + y + z)(x 2 + y 2 + z 2 xy yz xz)

24 Implicit Function Theorem II If the level surface (or implicit function) S : F(x, y, z) = c defines an explicit function z = z(x, y), and P S If F z (P) = 0, then z x = F x and z F z y = F y F z

25 Implicit Function Theorem II If the level surface (or implicit function) S : F(x, y, z) = c defines an explicit function z = z(x, y), and P S If F z (P) = 0, then z x = F x and z F z y = F y F z Remark In fact, Implicit Function Theorem tells us more If f z (a, b, c) = 0, then there exists a function z = g(x, y) defined on a small ball B(Q, δ) of Q(a, b) in xy-plane such that all the points on the level surface S : f (x, y, z) = f (a, b, c) is given by the graph of f, ie f (x, g(x, y)) = f (a, b, c) In other words, the level surface locally looks like a deformed sheet of xy-plane

26 Implicit Function Theorem II If the level surface (or implicit function) S : F(x, y, z) = c defines an explicit function z = z(x, y), and P S If F z (P) = 0, then z x = F x and z F z y = F y F z Remark In fact, Implicit Function Theorem tells us more If f z (a, b, c) = 0, then there exists a function z = g(x, y) defined on a small ball B(Q, δ) of Q(a, b) in xy-plane such that all the points on the level surface S : f (x, y, z) = f (a, b, c) is given by the graph of f, ie f (x, g(x, y)) = f (a, b, c) In other words, the level surface locally looks like a deformed sheet of xy-plane More generally, if f (a, b, c) is not a zero vector, then one can choose a right projection, so that locally the level surface S is a graph given by a differentiable function of 2 variables In particular, it is locally a smooth surface with a tangent plane This is why level surface can be thought as an implicit function, in which one can use it to define an implicit function in which one of the variable is expressed in terms of the other 2 variables

27 Example Suppose that the variables P, V and T satisfy the equation of famous ideal gas law: PV = nrt, where n and R are constants independent of P, V and T If P = P(V, T), V = V(P, T), and T = T(P, V) are implicit functions defined by means of the level surface S : PV = nrt (i) Evaluate the partial derivatives: P T (T, V), T V (ii) Simplify P T T V V P V (P, V) and (P, V) P Solution (i) As P = nrt P V, so T (T, V) = nr PV Similarly T = V nr, and hence T P nrt V (P, V) = And V = V nr P, so P (ii) P T T V V P = nr V (P, V) = nrt P 2 P nr nrt P 2 = nrt PV = 1

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

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 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

30 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

31 Find second order partial derivative y x 2 z(x, y) of the function z = z(x, y) defined implicitly by the equation 3x 2 + 2y 2 + z 2 + 8yz z + 8 = 0 Solution Replace the variable z in the implicit function F(x, y, z) = 3x 2 + 2y 2 + z 2 + 8yz z + 8 = 0 by an explicit function z = z(x, y), so it follows from the notation of the previous problem that z x = F x 6x z =, and F z 2z + 8y 1 y = F x 4y + 8z = F z 2z + 8y 1 So z xy = z x y = ( ) 6x y (2z + 8y 1) 2 6x = (2z + 8y 1) 2 (2z + 8y 1) y 6x = (2z + 8y 1) 2 (2z x + 8) 6x = (2z + 8y 1) 2 (8 12x 2z + 8y 1 )

32 Find all the points on the circle x 2 + y 2 = 1 at which the direction of fastest change of the function f (x, y) = x 2 + y 2 10x 8y is parallel to i + j A (1, 0) and (0, 1) B (1, 0) and ( 1, 0) C (0, 1) and ( 1, 0) D ( 1, 0) and (0, 1) E ( 1, 2 1 ) and ( 1, 1 ) Solution Answer is A The direction at which f changes fastest at P(a, b) is given by the direction f (a, b) So the question is to determine the point (x, y) with x 2 + y 2 = 1 such that f (x, y) = (2x 10, 2y 8) is parallel to (1, 1) So we have (2x 10, 2y 8) = 2λ(1, 1), ie (x, y) = (5 + λ, 4 + λ) Moreover, 1 = x 2 + y 2 = (5 + λ) 2 + (4 + λ) 2 = λ + 2λ 2, ie 0 = λ 2 + 9λ + 20 = (λ + 4)(λ + 5), so λ = 4 or 5 In this case, (x, y) = (1, 0) or (0, 1) respectively One can check that these two points satisfy the required conditions

33 Ẹvery function f(x, y) has at least one local maximum Solution False For example, f (x, y) = x does not If P is a critical point of f (x, y), then P is either a local maximum or a local minimum of f (x, y) Solution False For example, P can also be a saddle point Ẹvery function f (x, y) has a global minimum Solution False For example, f (x, y) = x does not The absolute maximum for a continuous function on a closed and bounded region must be attained at the boundary Solution False It can also be attained inside the region

34 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 )

35 Example The temperature in the vicinity of a polar bear is given by T(x, y, z) = e x + e 2y + e 3z If it is at (1, 1, 1), in what direction should the bear proceed in order to cool fastest? Solution

36 Example The temperature in the vicinity of a polar bear is given by T(x, y, z) = e x + e 2y + e 3z If it is at (1, 1, 1), in what direction should the bear proceed in order to cool fastest? Solution In order to cool the fastest, the bear should proceed in the direction in which T is decreasing the fastest; that is, in the direction T(1, 1, 1) As T(x, y, z) = T x i + T y j + T z k = ( e x, 2e 2y, 3e 3x ) Thus the required direction is T(1, 1, 1) = (1/e, 2/e 2, 3e 3 )

37 Example The temperature in the vicinity of a polar bear is given by T(x, y, z) = e x + e 2y + e 3z If it is at (1, 1, 1), in what direction should the bear proceed in order to cool fastest? Solution In order to cool the fastest, the bear should proceed in the direction in which T is decreasing the fastest; that is, in the direction T(1, 1, 1) As T(x, y, z) = T x i + T y j + T z k = ( e x, 2e 2y, 3e 3x ) Thus the required direction is T(1, 1, 1) = (1/e, 2/e 2, 3e 3 ) Example Let u = f (x, y, z) = e x2 sin(xy) In what direction from (1, π, 0) should one proceed to increase f most rapidly? Solution

38 Example The temperature in the vicinity of a polar bear is given by T(x, y, z) = e x + e 2y + e 3z If it is at (1, 1, 1), in what direction should the bear proceed in order to cool fastest? Solution In order to cool the fastest, the bear should proceed in the direction in which T is decreasing the fastest; that is, in the direction T(1, 1, 1) As T(x, y, z) = T x i + T y j + T z k = ( e x, 2e 2y, 3e 3x ) Thus the required direction is T(1, 1, 1) = (1/e, 2/e 2, 3e 3 ) Example Let u = f (x, y, z) = e x2 sin(xy) In what direction from (1, π, 0) should one proceed to increase f most rapidly? Solution f (x, y, z) = u x i + u y j + u x k = y cos(xy)e z2 i + x cos(xy)e z2 j 2z sin(xy)e z2 k, and hence f (1, π, 0) = ( π, 1, 0) So the required direction is the one given by ( π, 1, 0)

39 The maximum rate of change of function z = f (x, y) at the point P(3, 2) is A f (3, 2) B f (3, 2) C f (3, 2) D f (3, 2) E None of the above Answer is C Solution This is a a general fact, that the maximum rate of change of f (x 1,, x n ) at P(a 1, a 2,, a n ) is given by D n f (P), which is equal to f (p) along the direction n parallel to f (P)

40 Quiz: 10 minutes allowed 1 If z(x, y) = y + f (x 2 y 2 ) where f is a differentiable function of one variable on R, prove that y x z + x y z = x 2 Find the points on the surface S : xy 2 z 3 = 2 that are closest to the origin

41 Quiz: 10 minutes allowed 1 If z(x, y) = y + f (x 2 y 2 ) where f is a differentiable function of one variable on R, prove that y x z + x y z = x 2 Find the points on the surface S : xy 2 z 3 = 2 that are closest to the origin

42 Quiz Example If z(x, y) = y + f (x 2 y 2 ) where f is a differentiable function of one variable on R, prove that y z x + x y z = x

43 Quiz Example If z(x, y) = y + f (x 2 y 2 ) where f is a differentiable function of one variable on R, prove that y z x + x y z = x Solution This follows from chain rule that z x = x ( y + f (x2 y 2 ) ) = f (x 2 y 2 ) x (x2 y 2 ) = 2xf (x 2 y 2 ), and similarly, z y = y ( y + f (x2 y 2 ) ) = 1 + f (x 2 y 2 ) y (x2 y 2 ) = 1 2yf (x 2 y 2 ) Then we have y z x + x z y = y 2xf (x 2 y 2 ) + x (1 2yf (x 2 y 2 )) = x

44 Example Find the points on the surface S : xy 2 z 3 = 2 that are closest to the origin Solution I Let g(x, y, z) = log x + 2 log y + 3 log z = log(xy 2 z 3 ) = log 2 be constraint and d(x, y, z) = x 2 + y 2 + z 2 The Lagrangian equation is (2x, 2y, 2z) = d = λ g = λ( 1 x, 2 y, 3 z ), ie x2 = λ 2, y2 = λ and z 2 = 3λ 2 So 4 = 22 = (xy 2 z 3 ) 2 = x 2 y 4 z 6 = 33 λ 1+2+3, so λ 6 = , so λ = 2 Then d(x, y, z) = x 2 + y 2 + z 2 = ( )λ = 3λ = 2 3 ( ) So (x, y, z) = 1 2, 1, 3 2 is the point of S closest to origin Solution II Let x = f (y, z) = 2, for D = { (y, z) y z = 0 }, then the y 2 z 3 graph is exactly the surface S For any point P(f (y, z), y, z) in S, we define the squared distance function d(y, z) = x 2 + y 2 + z 2 = 4 + y 2 + z 2 = 4 + y2 y 4 z 6 y 4 z y2 2 + z2 3 + z2 3 + z y 4 z 6 y2 2 y2 2 z2 3 z2 3 z2 3 = ( 1 2 )2 ( 1 3 )3 = 6 = 2 3, 3 and equality holds if and only if 4 y 4 z 6 = y2 2 = z2 3 then the function d attains minimum value at P if and only if P is the closest to the origin

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