Several Variable Differential Calculus

Transcription

1 Several Variable Differential Calculus 1 Motivations The aim of studying the functions depending on several variables is to understand the functions which has several input variables and one or more output variables. For example, the following are Real valued functions of two variables x, y: (1) f(x, y) =x 2 + y 2 is a real valued function defined over IR 2. (2) f(x, y) = xy x 2 +y 2 is a real valued function defined over IR 2 \{(0, 0)} The real world problems like temperature distribution in a medium is a real valued function with more than 2 variables. The temperature function at time t and at point (x, y) has 3 variables. For example, the temperature distribution in a plate, (unit square) with zero temperature at the edges and initial temperature (at time t =0)T 0 (x, y) =sinπx sin πy, is T (t, x, y) =e π2kt sin πx sin πy. Another important problem of physics is Sound waves and water waves. The function u(x, t) = A sin(kx ωt) represents the traveling wave of the initial wave front sin kx. The Optimal cost functions, for example a manufacturing company wants to optimize the resources, for their produce, like man power, capital expenditure, raw materials etc. The cost function depends on these variables. Earning per share for Apple company ( ) has been modeled by z =0.379x 0.135y 3.45 where x is the sales and y is the share holders equity. 2 Limits Let IR 2 denote the set of all points (x, y) :x, y IR. The open ball of radius r with center (x 0,y 0 ) is denoted by B r ((x 0,y 0 )) = {(x, y) : (x x 0 ) 2 +(y y 0 ) 2 <r}. Definition: Apoint(a, b) is said to be interior point of a subset A of IR 2 if there exists r such that B r ((a, b)) A. AsubsetΩiscalledopenifeachpointofΩisaninteriorpoint of Ω. A subset is said to be closed in its compliment is an open subset of IR 2. For example (1). The open ball of radius δ: B δ ((0, 0)) = {(x, y) IR 2 : x 2 + y 2 < 1} is an open set 1

2 (2). Union of open balls is also an open set. (3). The closed ball of radius r : B r ((0, 0)) = {(x, y) : x 2 + y 2 r} is closed. A sequence {(x n,y n )} is said to converge to a point (x, y) inir 2 if for ɛ>0, there exists N IN such that xn x 2 + y n y 2 <ɛ, for all n N. Definition: Let Ω be a open set in IR 2,(a, b) Ω and let f be a real valued function defined on Ω except possibly at (a, b). Then the limit lim f(x, y) =L if for any ɛ>0 (x,y) (a,b) there exists δ>0 such that (x a)2 +(y b) 2 <δ = f(x, y) L <ɛ. Using the triangle inequality (as in one variable limit), one can show that if limit exists, then it is unique. That is, the limit is independent of choice path (x n,y n ) (a, b). 1. Examples (a) Consider the function f(x, y): f(x, y) = 4xy2. This function is defined in x 2 + y2 IR 2 \{(0, 0)}. Let ɛ>0. Then 4 xy 2 2(x 2 + y 2 )( x 2 + y 2 ). Therefore δ = ɛ For such δ, f(x, y) 0 <ɛ Figure 1: (b) Finding limit through polar coordinates Consider the function f(x, y) = x3 x 2 + y 2. This function is defined in IR2 \{(0, 0)}. 2

3 Taking x = r cos θ, y = r sin θ, weget f(r, θ) = r cos 3 θ r 0 as r 0. Figure 2: 2. Example of function which has different limits along different straight lines. Consider the function f(x, y) f(x, y) = xy. Then along the straight lines y = x 2 + y2 mx, wegetf(x, mx) = m. Hence limit does not exist. 1+m 2 Figure 3: 3. Example function where polar coordinates seem to give wrong conclusions Consider the function f(x, y) = 2x2 y x 4 + y. Taking the path, y = 2 mx2, we see that the 3

4 limit does not exist at (0, 0). Now taking x = r cos θ, y = r sin θ, we get f(r, θ) = 2r cos2 θ sin θ r 2 cos 4 θ +sin 2 θ. For any r>0, the denominator is > 0. Since cos 2 θ sin θ 1, we tend to think for Figure 4: a while that this limit goes to zero as r 0. So if we take the path r sin θ = r 2 cos 2 θ, (i.e., r = sin θ ), we get cos 2 θ f(r, θ) = 2sin2 θ 2sin 2 θ =1. 3 Continuity and Partial derivatives Let f be a real valued function defined in a ball around (a, b). Then Definition: f is said to be continuous at (a, b) if lim f(x, y) =f(a, b) (x,y) (a,b) 4

5 Example: The function xy x 2 + y 2 0 x f(x, y) = 2 +y 2 0 x = y =0 Let ɛ>0. Then f(x, y) 0 = x y x 2 +y 2 x. Soifwechooseδ = ɛ, then f(x, y) ɛ. Figure 5: Therefore, f is continuous at (0, 0). Partial Derivatives: The partial derivative of f with respect to x at (a, b) is defined as f 1 (a, b) = lim (f(a + h, b) f(a, b)). x h 0 h similarly, the partial derivative with respect to y at (a, b) is defined as Example: Consider the function f 1 (a, b) = lim (f(a, b + k) f(a, b)). y k 0 k xy (x, y) (0, 0) x f(x, y) = 2 +y 2 0 (x, y) (0, 0) 5

6 As noted earlier, this is not a continuous function, but f(h, 0) f(0, 0) 0 0 f x (0, 0) = lim = lim =0. h 0 h h 0 h Similarly, we can show that f y (0, 0) exists. Also for a continuous function, partial derivatives need not exist. For example f(x, y) = Figure 6: x + y. This is a continuous function at (0, 0). Indeed, for any ɛ>0, we can take δ<ɛ/2. But partial derivatives do not exist at (0, 0) Sufficient condition for continuity: Theorem: Suppose one of the partial derivatives exist at (a, b) and the other partial derivative is bounded in a neighborhood of (a, b). Then f(x, y) is continuous at (a, b). Proof: Let f y exists at (a, b). Then f(a, b + k) f(a, b) =kf y (a, b)+ɛ 1 k, where ɛ 1 0ask 0. Since f x exists and bounded in a neighborhood of at (a, b), f(a + h, b + k) f(a, b) =f(a + h, b + k) f(a, b + k)+f(a, b + k) f(a, b) =hf x (a + θh, b + k)+kf y (a, b)+ɛ 1 k hm + k f y (a, b) + ɛ 1 k 0 as h, k 0. 6

7 Directional derivatives, Definition and examples Let ˆp = p 1 î + p 2 ĵ be any unit vector. Then the directional derivative of f(x, y) at(a, b) in the direction of ˆp is f(a + sp 1,b+ sp 2 ) f(a, b) Dˆp f(a, b) = lim. s 0 s Example f(x, y) =x 2 + xy at P (1, 2) in the direction of unit vector u = 1 2 î ĵ. Dˆp f(1, 2) = lim f(1 + s 2, 2+ s s 0 1 = lim s 0 s 2 ) f(1, 2) ( s s 2 + s( ) 2 ) = The existence of partial derivatives does not guarantee the existence of directional derivatives in all directions. For example take xy x 2 + y 2 0 x f(x, y) = 2 +y 2 0 x = y =0. Let p =(p 1,p 2 ) such that p p 2 2 = 1. Then the directional derivative along p is f(hp 1,hp 2 ) f(0, 0) p 1 p 2 D p f(0, 0) = lim = lim h 0 h h 0 h(p p 2 2) exist if and only if p 1 =0orp 2 =0. The existence of all directional derivatives does not guarantee the continuity of the function. For example take x 2 y (x, y) (0, 0) x f(x, y) = 4 +y 2. 0 x = y =0 7

8 Let p =(p 1,p 2 ) such that p p 2 2 = 1. Then the directional derivative along p is f(sp 1,sp 2 ) f(0, 0) D p f(0, 0) = lim s 0 s s 3 p 2 = lim 1p 2 s 0 s(s 4 p s 2 p 2 2) = p2 1p 2 p 2 2 if p 2 0 In case of p 2 =0, we can compute the partial derivative w.r.t y to be 0. Therefore all the directional derivatives exist. But this function is not continuous (y = mx 2 and x 0). 4 Differentiability Let D be an open subset of IR 2.Then Definition: A function f(x, y) :D IR is differentiable at a point (a, b) ofd if there exists ɛ 1 = ɛ(h, k),ɛ 2 = ɛ 2 (h, k) such that f(a + hb + k) f(a, b) =hf x (a, b)+kf y (a, b)+hɛ 1 + kɛ 2, where ɛ 1,ɛ 2 0as(h, k) (0, 0). Problem: Show that the following function f(x, y) is is not differentiable at (0, 0) x sin 1 f(x, y) = + y sin 1, xy 0 x y 0 xy =0 Solution: f(x, y) x + y 2 x 2 + y 2 implies that f is continuous at (0, 0). Also f(h, 0) f(0, 0) f x (0, 0) = lim =0. h 0 h f(0,k) f(0, 0) f y (0,k) = lim =0. k 0 k If f is differentiable, then there exists ɛ 1,ɛ 2 such that f(h, k) f(0, 0) = ɛ 1 h + ɛ 2 k 8

9 Figure 7: where ɛ 1,ɛ 2 0ash, k 0. Now taking h = k, weget f(h, h) =(ɛ 1 + ɛ 2 )h = 2h sin 1 h = h(ɛ 1 + ɛ 2 ). So as h 0, we get sin 1 0, a contradiction. h Problem: Show that the function f(x, y) = xy is not differentiable at the origin. Solution: Easy to check the continuity (take δ = ɛ). 0 0 f x (0, 0) = lim =0, and similar calculation shows f y (0.0) = 0 h 0 h So if f is differentiable at (0, 0), then there exist, ɛ 1,ɛ 2 such that f(h, k) =ɛ 1 h + ɛ 2 k. Taking h = k, we get h =(ɛ 1 + ɛ 2 )h. This implies that (ɛ 1 + ɛ 2 ) 0. Notations: 1. Δf = f(a + h, b + k) f(a, b), the total variation of f 2. df = hf x (a, b)+kf y (a, b), the total differential of f. 3. ρ = h 2 + k 2 9

10 Equivalent condition for differentiability: Δf df Theorem: f is differentiable at (a, b) lim =0. ρ 0 ρ Proof. Suppose f(x, y) is differentiable. Then, there exists ɛ 1,ɛ 2 such that f(a + h, b + k) f(a, b) =hf x (a, b)+kf y (a, b)+hɛ 1 + kɛ 2, where ɛ 1,ɛ 2 0as(h, k) (0, 0). Therefore, we can write Δf df = ɛ 1 ( h ρ ρ )+ɛ 2( k ρ ), where df = hf x (a, b)+kf y (a, b) andρ = h 2 + k 2. Now since h 1, k 1weget ρ ρ Δf df lim ρ 0 ρ =0. On the other hand, if lim ρ 0 Δf df ρ =0,then Δf = df + ɛρ, ɛ 0 as h, k 0. ɛρ = = ɛρ h + k h + ɛρ h + k k ɛρ sgn(h) ɛρ sgn(k) h + h + k h + k k Therefore, we can take ɛ 1 = ɛρ sgn(h) h + k and ɛ 2 = ɛρ sgn(k) h + k. x 2 y 2 (x, y) (0, 0) x Example: Consider the function f(x, y) = 2 +y 2. 0 x = y =0 Partial derivatives exist at (0, 0) and f x (0, 0) = 0,f y (0, 0) = 0. By taking h = ρ cos θ, k = Figure 8: 10

11 ρ sin θ, weget Therefore, Δf df ρ Δf df ρ = h2 k 2 ρ 3 = ρ4 cos 2 θ sin θ ρ 3 = ρ cos 2 θ sin θ. ρ 0asρ 0. Therefore f is differentiable at (0, 0). x 2 y (x, y) 0 x Example: Consider f(x, y) = 2 +y 2 0 x = y =0. Partial derivatives exist at (0, 0) and f x (0, 0) = f y (, 0) = 0. By taking h = ρ cos θ, k = ρ sin θ, weget Δf df = h2 k = ρ3 cos 2 θ sin θ =cos 2 θ sin θ. ρ ρ 3 ρ 3 The limit does not exist. Therefore f is NOT differentiable at (0, 0). The following theorem is on Sufficient condition for differentiability: Theorem: Suppose f x (x, y) andf y (x, y) exist in an open neighborhood containing (a, b) and both functions are continuous at (a, b). Then f is differentiable at (a, b). Proof. Since f is continuous at (a, b), there exists a neighborhood N(say) of (a, b) at y every point of which f y exists. Wetake(a + h, b + k), a point of this neighborhood so that (a + h, b), (a, b + k) also belongs to N. We write f(a+h, b+k) f(a, b) = {f(a+h, b+k) f(a+h, b)} + {f(a+h, b) f(a, b)}. Consider a function of one variable φ(y) = f(a + h, y). Since f y exists in N,φ(y) is differentiable with respect to y in the closed interval [b, b + k] and as such we can apply Lagrange s Mean Value Theorem, for function of one variable y in this interval and thus obtain φ(b + k) φ(b) = kφ (b + θk), 0 <θ<1 = kf y (a + h, b + θk) f(a + h, b + k) f(a + h, b) = kf y (a + h, b + θk), 0 <θ<1. 11

12 Now, if we write f y (a + h, b + θk) f y (a, b) = ɛ 2 (a function of h, k) then from the fact that f y is continuous at (a,b). we may obtain ɛ 2 0 as (h, k) (0, 0). Again because f x exists at (a, b) implies f(a + h, b) f(a, b) = hf x (a, b) + ɛ 1 h, where ɛ 1 0ash 0. Combining all these we get f(a + h, b + k) f(a, b) = k[f y (a, b) + ɛ 2 ] + hf x (a, b) + ɛ 1 h = hf x (a, b) + kf y (a, b) + ɛ 1 h + ɛ 2 k where ɛ 1,ɛ 2 are functions of (h, k) and they tend to zero as (h, k) (0, 0). This proves that f(x, y) is differentiable at (a, b). Remark: The above proof still holds if f y is continuous and f x exists at (a, b). There are functions which are Differentiable but the partial derivatives need not be continuous. For example, consider the function x 3 sin 1 + y 3 sin 1 xy 0 x f(x, y) = 2 y 2 0 xy =0. Then 3x 2 sin 1 2cos 1 xy 0 x f x (x, y) = 2 x 2 0 xy =0 Also f x (0, 0) = lim h 0 f(h,0) f(0,0) h =0. So partial derivatives are not continuous at (0, 0). f(δx, Δy) =(Δx) 3 sin 1 1 (Δx) 2 +(Δy)3 sin (Δy) 2 =0+0+ɛ 1 Δx + ɛ 2 Δy 12

13 Figure 9: where ɛ 1 =(Δx) 2 sin 1 (Δx) 2 and ɛ 2 =(Δy) 2 sin 1 (Δy) 2. It is easy to check that ɛ 1,ɛ 2 0. So f is differentiable at (0, 0). There are functions for which directional derivatives exist in any direction, but the function is not differentiable. Example: Consider the function y f(x, y) = y x2 + y 2 y 0 0 y =0 (Exercise problem) Chain rule: Partial derivatives of composite functions: Let z = F (u, v) andu = φ(x, y),v = ψ(x, y). Then z = F (φ(x, y),ψ(x, y)) as a function of x, y. Suppose F, φ, ψ have continuous partial derivatives, then we can find the partial derivatives of z w.r.t x, y as follows: Let x be increased by Δx, keeping y constant. Then the increment in u is Δ x u = u(x +Δx, y) u(x, y) and similarly for v. Then the increment in z is (as z is differentiable as a function of u, v ) Δ x z := z(x +Δx, y +Δy) z(x, y) = F u Δ xu + F v Δ xv + ɛ 1 Δ x u + ɛ 2 Δ x v Now dividing by Δx Δ x z Δx = F Δ x u u Δx + F Δ x v v Δx + ɛ Δ x u 1 Δx + ɛ Δ x v 2 Δx 13

14 Taking Δx 0, we get z x = F u u x + F v v x + ( lim ɛ 1) u Δx 0 x + ( lim ɛ 2) v Δx 0 x u x + F v v x = F u similarly, one can show z y = F u u y + F v v y. Example: Let z = ln(u 2 + v 2 ),u= e x+y2,v = x 2 + y. Then z u = 2u,z u 2 +v v = 1,u u 2 +v x = e x+y2,v x =2x. Then z x = 2u u 2 + v ex+y2 + 2x u 2 + v Derivative of Implicitly defined function Let y = y(x) be defined as F (x, y) =0, where F, F x,f y are continuous at (x 0,y 0 )and F y (x 0,y 0 ) 0. Then dy = Fx dx F y at (x 0,y 0 ). Increase x by Δx, theny receives Δy increment and F (x +Δx, y +Δy) =0. Also where ɛ 1,ɛ 2 0asΔx 0. This is same as 0=ΔF = F x Δx + F y Δy + ɛ 1 Δx + ɛ 2 Δy Δy Δx = F x + ɛ 1 F y + ɛ 2 Now taking limit Δx 0, we get dy dx = Fx F y. Example: The function y = y(x) ase y e x + xy =0. ThenF x = e x + y, F y = e y + x. Then dy = ex y. dx e y +x Proposition If f(x, y) is differentiable, then the directional derivative in the direction ˆp at (a, b)is Dˆp f(a, b) = f(a, b) ˆp. Proof: Let ˆp =(p 1,p 2 ). Then from the definition, f(a + sp 1,b+ sp 2 ) f(a, b) f(x(s),y(s)) f(x(0),y(0)) lim = lim s 0 s s 0 s 14

15 where x(s) =a + sp 1,y(s) =b + sp 2. From the chain rule, f(x(s),y(s)) f(x(0),y(0)) lim s 0 s = f (a, b)dx x ds + f y (a, b)dy ds = f(a, b) (p 1,p 2 ). Using the directional derivatives, we can also find the Direction of maximum rate of change. Properties of directional derivative for differentiable function Dˆp f = f ˆp = f cos θ. So the function f increases most rapidly when cos θ =1or when ˆp is the direction of f. The derivative in the direction f is equal to f. f Similarly, f decreases most rapidly in the direction of f. The derivative in this direction is Dˆp f = f. Finally, the direction of no change is when θ = π. i.,e., ˆp f. 2 Problem: Find the direction in which f = x 2 /2+y 2 /2 increases and decreases most rapidly at the point (1, 1). Also find the direction of zero change at (1, 1). Example Functions for which all directional derivatives exist but not differentiable. Consider the function y f(x, y) = y x2 + y 2, y 0 0 y =0 f(sp 1,sp 2 ) f(0, 0) Dˆp f(a, b) =lim = p 2 s 0 s p f x (0, 0) = lim =0, f y (0, 0) = lim h 0 h k 0 k In this case, both Dˆp f(a, b) and f(a, b) ˆp exists but not equal. k k k 2 =1. Tangents and Normal to Level curves Let f(x, y) be differentiable and consider the level curve f(x, y) =c. Let r (t) =g(t)î + h(t)ĵ be its parametrization. for example f(x, y) =x 2 + y 2 has x(t) =a cos t, y(t) =a sin t as level curve x 2 + y 2 = a 2, which is a circle of radius a. Now differentiating the equation f(x(t),y(t)) = a 2 with respect to t, weget dx f x dt + f dy y dt =0. Now since r (t) =x (t)î + y (t)ĵ is the tangent to the curve, we can infer from the above equation that f is the direction of Normal. Hence we have Equation of Normal at (a, b) is x = a + f x (a, b)t, y = b + f y (a, b)t, t IR Equation of Tangent is (x a)f x (a, b)+(y b)f y (a, b) =0. 15

16 Example Find the normal and tangent to x2 4 + y2 =2at( 2, 1). Solution: We find f = x 2î +2yĵ ( 2,1) = î +2ĵ. Therefore, the Tangent line through Figure 10: ( 2, 1) is (x +2)+2(y 1) = 0. Tangent Plane and Normal lines Let r (t) =g(t)î + h(t)ĵ + k(t)ˆk is a smooth level curve(space curve) of the level surface f(x, y, z) =c. Then differentiating f(x(t),y(t),z(t)) = c with respect to t and applying chainrule,weget f(a, b, c) (x (t),y (t),z (t)) = 0 Now as in the above, we infer the following: Normal line at (a, b, c) isx = a + f x t, y = b + f y t, z = c + f z t. Tangent plane: (x a)f x +(y b)f y +(z c)f z =0. Example: Find the tangent plane and normal line of f(x, y, z) =x 2 + y 2 + z 9=0at (1, 2, 4). f +2xî+2yĵ+ˆk at (1,2,4) is = 2î+4ĵ+ˆk and tangent plane is 2(x 1)+4(y 2)(z 4) = 0. Figure 11: Then normal line is x =1+2t, y =2+4t, z =4+t. 16

17 Problem: Find the tangent line to the curve of intersection of two surfaces f(x, y, z) = x 2 + y 2 2=0,z R, g(x, y, z) =x + z 4=0. Solution: The intersection of these two surfaces is an an ellipse on the plane g =0. Figure 12: The direction of normal to g(x, y, z) =0at(1, 1, 3) is î + ˆk and normal to f(x, y, z) =0is 2î +2ĵ. The required tangent line is orthogonal to both these normals. So the direction of tangent is v = f g =2î 2ĵ 2ˆk. Tangent through (1, 1, 3) is x =1+2t, y =1 2t, z =3 2t. Linearization: Let f be a differentiable function in a rectangle containing (a, b). The linearization of a function f(x, y) atapoint(a, b) wheref is differentiable is the function L(x, y) =f(a, b)+f x (a, b)(x a)+f y (a, b)(y b). The approximation f(x, y) L(x, y) is the standard linear approximation of f(x, y) at (a, b). Since the function is differentiable, E(x, y) =f(x, y) L(x, y) =ɛ 1 (x a)+ɛ 2 (y b) where ɛ 1,ɛ 2 0asx a, y b. The error of this approximation is E(x, y) M 2 ( x a + y b )2. where M =max{ f xx, f xy, f yy }. Example: Find the linearization and error in the approximation of f(x, y, z) =x 2 xy y2 +3at(3, 2). 17

18 f(3, 2) = 8,f x (3, 2) = 4,f y (3, 2) 1. So L(x, y) =8+4(x 3) (y 2) = 4x y 2 Also max{ f xx, f xy, f yy } =2and E(x, y) ( x 3 + y 2 ) 2 If we take the rectangle R : x 3 0.1, y Taylor s theorem Higher order mixed derivatives: It is not always true that the second order mixed derivatives f xy = ( f )andf x y yx = ( f ) y x are equal. The following is the example xy(x 2 y 2 ), x 0,y 0 x Example: f(x, y) = 2 +y 2. 0 x = y =0 Then Figure 13: f(h, k) f(h, 0) 1 hk(h 2 k 2 ) f y (h, 0) = lim = lim = h k 0 k k 0 k h 2 + k 2 18

19 Also f y (0, 0) = 0. Therefore, f y (h, 0) f y (0, 0) f xy (0, 0) = lim h 0 h h 0 = lim =1 h 0 h Now f(h, k) f(0,k) 1 hk(h 2 k 2 ) f x (0,k) = lim = lim = k h 0 h h 0 h h 2 + k 2 and f x (0,k) f x (0, 0) f yx (0, 0) = lim = 1. h 0 k /// The following theorem is on the sufficient condition for equality of mixed derivatives. We omit the proof. Theorem: If f,f x,f y,f xy,f yx are continuous in a neighbourhood of (a, b). Then f xy (a, b) = f yx (a, b). But this is not a necessary condition as can be seen from the following example Example: f xy,f yx not continuous but mixed derivatives are equal. x 2 y 2 x 0,y 0 x Consider the function f(x, y) = 2 +y 2. 0 x = y =0 Here f xy (0, 0) = f yx (0, 0) but they are not continuous at (0, 0) (Try!). Taylor s Theorem: 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) + 1 2! (h2 f xx +2hkf xy + k 2 f yy ) (a,b) + 1 3! (h3 f xxx +3h 2 kf xxy +3hk 2 f xyy + k 3 f yyy ) (a,b) n! (h x + k 1 y )n f (a,b) + (n +1)! (h x + k y )n+1 f (a+ch,b+ck) where (a + ch, b + ck) is a point on the line segment joining (a, b) and(a + h, b + k). Proof. Proof follows by applying the Taylor s theorem, Chain rule on the one dimensional function φ(t) =f(x + ht, y + kt) att =0. The function f(x, y) = xy does not satisfy the hypothesis of Taylor s theorem around 19

20 (0, 0). Indeed, the first order partial derivatives are not continuous at (0, 0). Error estimation: problem: The function f(x, y) =x 2 xy + y 2 is approximated by a first degree Taylor s polynomial about (2, 3). Find a square x 2 <δ, y 3 <δsuch that the error of approximation is less than or equal to 0.1. Solution: We have f x =2x y, f y =2y x, f xx = 1,f xy = 1,f yy = 2. The maximum Figure 14: error in the first degree approximation is R B ( x 2 + y 3 )2 2 where B =max{ f xx, f xy, f yy } =max2, 1, 2=2. Therefore, we want to determine δ such that R 2 2 (δ + δ)2 < 0.1 or 4δ 2 < 0.1, or δ< Maxima and Minima Derivative test: From our discussion on one variable calculus, it is enough to determine the sign of Δf = f(a +Δx, b +Δy) f(a, b). Now by Taylor s theorem, Δf = 1 ( fxx (a, b)(δx) 2 +2f xy (a, b)δxδy + f yy (a, b)(δy) 2) + α(δρ)

21 where Δρ = (Δx) 2 +(Δy) 2. We use the notation A = f xx,b = f xy,c = f yy In polar form, Δx =Δρ cos φ, Δy =Δρ sin φ Then we have Suppose A 0,then Δf = 1 2 (Δρ)2 ( A cos 2 φ +2B cos φ sin φ + C sin 2 φ +2αΔρ ). Δf = 1 2 (Δρ)2 ( (A cos φ + B sin φ) 2 +(AC B 2 )sin 2 φ A ) +2αΔρ Now we consider the 4 possible cases: Case 1: Let AC B 2 > 0,A<0. Then (A cos φ + B sin φ) 2 0, sin 2 φ 0 implies Δf = 1 2 (Δρ)2 ( m 2 +2αΔρ ) where m is independent of Δρ, αδρ 0asΔρ 0. Hence for Δρ small, Δf <0. Hence the point (a, b) isapointoflocalmaximum Case 2: Let AC B 2 > 0,A>0. In this case Δf = 1 2 (Δρ)2 (m 2 + αδρ) So Δf >0. Hence the point (a, b) is a point of local minimum. Case 3: 1. Let AC B 2 < 0,A>0. When we move along φ =0, we have Δf = 1 2 (Δρ)2 (A +2αΔρ) > 0. for Δρ small. When we move along tan φ 0 = A/B, then Δf = 1 2 (Δρ)2 ( AC B 2 A ) sin 2 φ 0 +2αΔρ < 0 for Δρ small. so we don t have constant sign along all directions. Hence (a, b) is 21

22 neither a point of maximum nor a point of minimum. Saddle point. Such point (a, b) is called 2. Let AC B 2 < 0,A<0. Similar as above the sign along path φ =0is< 0andtanφ 0 = A/B is > Let AC B 2 < 0,A=0. In this case B 0and Δf = 1 2 (Δρ)2 (sin φ(2b cos φ + C sin φ)+2αδρ) for small φ, 2B cos φ + C sin φ is close to 2B, but sin φ changes sign for φ>0or φ<0. Again here (a, b) is a saddle point. Case 4: Let AC B 2 =0. Again in this case it is difficult to decide the sign of Δf. For instance, if A 0, Δf = 1 2 (Δρ)2 ( (A cos φ + B sin φ) 2 A ) +2αΔρ When φ = arctan( A/B), the sign of Δf is determined by the sign of α. So Additional investigation is required. No conclusion can be made with AC B 2 =0. We summarize the derivative test in two variables in the following table: S.No. Condition Nature 1 AC B 2 > 0,A>0 local minimum 2 AC B 2 > 0,A<0 local maximum 3 AC B 2 < 0 Saddle point 4 AC B 2 =0 No conclusion Example: Find critical points and their nature of f(x, y) =xy x 2 y 2 2x 2y +4 f x = y 2x 2=0, f y = x 2y 2=0 Therefore, the point ( 2, 2) is the only critical point. Also f xx = 2, f yy = 2, f xy =1. Therefore, AC B 2 =3> 0andA = 2 < 0. Therefore, ( 2, 2) is a point of local 22

23 Figure 15: maximum. The following is an example where the derivative test fails. Example: Consider the function f(x, y) =(x y) 2,thenf x =0,f y = 0 implies x = y. Also, AC B 2 = 0. Moreover, all third order partial derivatives are zero. so no further information can be expected from Taylor s theorem. Global/Absolute maxima and Minima on closed and bounded domains: 1. Find all critical points of f(x, y). These are the interior points where partial derivatives can be defined. 2. Restrict the function to the each piece of the boundary. This will be one variable function defined on closed interval I(say) and use the derivative test of one variable calculus to find the critical points that lie in the open interval and their nature. 3. Find the end points of these intervals I and evaluate f(x, y) at these points. 4. The global/absolute maximum will be the maximum of f among all these points. 5. Similarly for global minimum. Example: Find the absolute maxima and minima of f(x, y) =2+2x +2y x 2 y 2 on the triangular region in the first quadrant bounded by the lines x =0,y =0,y =9 x. Solution: f x =2 2x =0,f y =2 2y = 0 implies that x =1,y = 1 is the only critical point and f(1, 1) = 4. f xx = 2,f yy = 2,f xy = 0. Therefore, AC B 2 =4> 0and A<0. So this is local maximum. 23

24 case 1: On the segment y =0,f(x, y) =f(x, 0) = 2 + 2x x 2 defined on I =[0, 9]. f(0, 0) = 2,f(9, 0) = 61 and the at the interior points where f (x, 0) = 2 2x =0. So x = 1 is the only critical point and f(1, 0) = 3. case 2: On the segment x =0,f(0,y)=2+2y y 2 and f y =2 2y = 0 implies y =1 and f(0, 1) = 3. Case 3: On the segment y =9 x, wehave f(x, 9 x) = 61+18x 2x 2 and the critical point is x =9/2. At this point f(9/2, 9/2) = 41/2. finally, f(0, 0) = 2,f(9, 0) = f(0, 9) = 61. so the global maximum is 4 at (1, 1) and minimum is 61 at (9, 0) and (0, 9). 7 Constrained minimization First, we describe the substitution method. Consider the problem of finding the shortest distance from origin to the plane z =2x + y 5. Here we minimize the function f(x, y, z) =x 2 + y 2 + z 2 subject to the constraint 2x + y z 5=0. Substituting the constraint in the function, we get h(x, y) =f(x, y, 2x + y 5) = x 2 + y 2 +(2x + y 5) 2. The critical points of this function are h x =2x +2(2x + y 5)(2) = 0, h y =2y +2(2x + y 5) = 0. This leads to x =5/3,y =5/6. Then z =2x + y 5 implies z = 5/6. We can check that AC B 2 > 0andA>0. So the point (5/3, 5/6, 5/6) is a point of minimum. Does this substitution method always work? The answer is NO. The following example explains Example: The shortest distance from origin to x 2 z 2 =1. This is minimizing f(x, y, z) =x 2 + y 2 + z 2 subject to the constraint x 2 z 2 =1. Then substituting z 2 = x 2 1inf, weget h(x, y) =f(x, y, x 2 1) = 2x 2 + y

25 The critical points of the function are h x =4x =0,h y =2y =0. Thatis,x =0,y = 0,z 2 = 1. But this point is not on the hyperbolic cylinder. To overcome this difficulty, we can substitute x 2 = z 2 +1inf and find that z = y =0andx = ±1. These points are on the hyperbolic cylinder and we can check that AC B 2 > 0,A > 0. This implies the points are of local minimum nature. In the substitution method, once we substitute the constraint in the minimizing function, then the the domain of the function will be the domain of the minimizing function. Then the critical points can belong to this domain which may not be the domain of constraints. This is overcome by the following: Lagrange Multiplier Method: Imagine a small sphere centered at the origin. Keep increasing the radius of the sphere untill the sphere touches the hyperbolic cylinder. The required smallest distance is the radius of that sphere which touches the cylinder. When the sphere touches the cylinder, both these surfaces has common tangent plane. So at the point of touching, both surfaces has normal proportional. That is f = λ g for some λ. Now solving this equations along with g =0givesthe points of extrema. In the above example, taking f = x 2 + y 2 + z 2 and g = x 2 z 2 1=0, we get 2xî +2yĵ +2z k = λ(2xî 2z k) This implies, 2x =2λx, 2y =0 2z = 2λz. x = 0 does not satisfy g =0. So from first equation we get λ =1. Then2z = 2z. Thatisz =0, and y = 0. Therefore, the critical points are (x, 0, 0). Substituting this in the constraint equation, we get x = ±1. Hence the points of extrema are (±, 0, 0). Caution: The Method of Lagrange multiplies gives only the points of extremum. To find the maxima or minima one has to compare the function values at these extremum points. Method of Lagrange multipliers with many constraints in n variables 1. Number of constraints m should be less than the number of independent variables n say g 1 =0,g 2 =0,...g m =0. 2. Write the Lagrange multiplier equation: f = m λ i g i. i=1 25

26 (a) Lagrange multipliers Figure 16: 3. Solve the set of m + n equations to find the extremal points f = m λ i g i, i=1 g i =0,i=1, 2,...m 4. Once we have extremum points, compare the values of f at these points to determine the maxima and minima. Refereces: 1. Thomas Calculus Chapter N. Piskunov, Differential and Integral calculus 26