18.02SC Supplementary Problem Solutions: Partial Differentiation

2. Partial Differentiation 2A. Functions and Partial Derivatives 2A-1 In the pictures below, not all of the level curves are labeled. In (c) and (d), the picture is the same, but the labelings are different. In more detail: b) the origin is the level curve 0; the other two unlabeled level curves are .5 and 1.5; c) on the left, two level curves are labeled; the unlabeled ones are 2 and 3; the origin is the level curve 0; d) on the right, two level curves are labeled; the unlabeled ones are −1 and −2; the origin is the level curve 1; The crude sketches of the graph in the first octant are at the right. 2 1 4 1 1 0 -3 2 1 0 0 0 3 2 -2 -1 b b) zx = c) fx = 3 cos(3x + 2y), fy = 2 cos(3x + 2y) 2 d) fx = 2xyex y , fy = x2 ex 2 y e) zx = ln(2x + y) + 2 c e c, d 2A-2 a) fx = 3x2 y − 3y 2 , fy = x3 − 6xy + 4y x 1 , zy = − 2 y y 2x , 2x + y zy = d x 2x + y f) fx = 2xz, fy = −2z 3 , fz = x2 − 6yz 2 2A-3 a) both sides are mnxm−1 y n−1 x−y −x y b) fx = , fxy = (fx )y = ; fy = , 2 3 (x + y) (x + y) (x + y)2 c) fx = −2x sin(x2 + y), fxy = (fx )y = −2x cos(x2 + y); fy = − sin(x2 + y), fyx = − cos(x2 + y) · 2x. d) both sides are f ′ (x)g ′ (y). fyx = −(y − x) . (x + y)3 2A-4 (fx )y = ax + 6y, (fy )x = 2x + 6y; therefore fxy = fyx ⇔ a = 2. By inspection, one sees that if a = 2, f (x, y) = x2 y + 3xy 2 is a function with the given fx and fy . 2A-5 a) wx = aeax sin ay, wy = eax a cos ay, wxx = a2 eax sin ay; wyy = eax a2 (− sin ay); b -1 -2 1 a 1 a therefore wyy = −wxx . 2x 2(y 2 − x2 ) , w = . If we interchange x and y, the function xx x2 + y 2 (x2 + y 2 )2 2 2 w = ln(x + y ) remains the same, while wxx gets turned into wyy ; since the interchange just changes the sign of the right hand side, it follows that wyy = −wxx . b) We have wx = 2B. Tangent Plane; Linear Approximation 2B-1 a) zx = y 2 , zy = 2xy; therefore at (1,1,1), we get zx = 1, tangent plane is z = 1 + (x − 1) + 2(y − 1), or z = x + 2y − 2. 0 zy = 2, so that the 2. PARTIAL DIFFERENTIATION 1 b) wx = −y 2 /x2 , wy = 2y/x; therefore at (1,2,4), we get wx = −4, wy = 4, so that the tangent plane is w = 4 − 4(x − 1) + 4(y − 2), or w = −4x + 4y. 2B-2 a) zx = � x x2 y2 = x ; z by symmetry (interchanging x and y), zy = y ; then the z + x0 y0 x0 y0 tangent plane is z = z0 + (x − x0 ) + (y − y0 ), or z = x + y , since x20 + y02 = z02 . z0 z0 z0 z0 b) The line is x = x0 t, y = y0 t, z = z0 t; substituting into the equations of the cone and the tangent plane, both are satisfied for all values of t; this shows the line lies on both the cone and tangent plane (this can also be seen geometrically). 2B-3 Letting x, y, z be respectively the lengths of the two legs and the hypotenuse, we � have z = x2 + y 2 ; thus the calculation of partial derivatives is the same as in 2B-2, and 3 4 7 we get Δz ≈ Δx + Δy. Taking Δx = Δy = .01, we get Δz ≈ (.01) = .014. 5 5 5 R1 R2 . From this we calculate 2B-4 From the formula, we get R = R1 + R2 � �2 � �2 ∂R R2 ∂R R1 = , and by symmetry, = . ∂R1 R1 + R2 ∂R2 R1 + R2 1 4 Substituting R1 = 1, R2 = 2 the approximation formula then gives ΔR = ΔR1 + ΔR2 . 9 9 4 1 5 By hypothesis, |ΔRi | ≤ .1, for i = 1, 2, so that |ΔR| ≤ (.1) + (.1) = (.1) ≈ .06; thus 9 9 9 2 R = = .67 ± .06. 3 2B-5 a) We have f (x, y) = (x+y +2)2 , fx = 2(x+y +2), at (0, 0), fx (0, 0) = fy (0, 0) = 4, f (0, 0) = 4; fy = 2(x+y +2). Therefore linearization is 4 + 4x + 4y; at (1, 2), fx (1, 2) = fy (1, 2) = 10, f (1, 2) = 25; linearization is 10(x − 1) + 10(y − 2) + 25, or 10x + 10y − 5. b) f = ex cos y; fx = ex cos y; fy = −ex sin y . linearization at (0, 0): 1 + x; linearization at (0, π/2): −(y − π/2) � � � � ∂V ∂V ∂V ∂V 2 2 = 2πrh, = πr ; ΔV ≈ Δr + Δh. 2B-6 We have V = πr h, ∂r ∂h ∂r 0 ∂h 0 Evaluating the partials at r = 2, h = 3, we get ΔV ≈ 12πΔr + 4πΔh. Assuming the same accuracy |Δr| ≤ ǫ, |Δh| ≤ ǫ for both measurements, we get 1 |ΔV | ≤ 12π ǫ + 4π ǫ = 16π ǫ, which is < .1 if ǫ < < .002 . 160π � y ∂r x ∂r y 2B-7 We have r = x2 + y 2 , θ = tan−1 ; = , = . x ∂x r ∂y r Therefore at (3, 4), r = 5, and Δr ≈ 35 Δx + 54 Δy. If |Δx| and |Δy| are both ≤ .01, then |Δr| ≤ Similarly, ∂θ −y ; = 2 ∂x x + y2 3 5 |Δx| + 45 |Δy| = 7 5 (.01) = .014 (or .02). ∂θ x , so at the point (3, 4), = 2 ∂y x + y2 2 S. 18.02 SOLUTIONS TO EXERCISES 3 |Δθ| ≤ | −4 25 Δx| + | 25 Δy | ≤ 7 25 (.01) = .0028 (or .003). Since at (3, 4) we have |ry | > |rx |, r is more sensitive there to changes in y; by analogous reasoning, θ is more sensitive there to x. 2B-9 a) w = x2 (y + 1); wx = 2x(y + 1) = 2 at (1, 0), and wy = x2 = 1 at (1, 0); therefore w is more sensitive to changes in x around this point. b) To first order approximation, Δw ≈ 2Δx + Δy, using the above values of the partial derivatives. If we want Δw = 0, then by the above, 2Δx + Δy = 0, or Δy/Δx = −2 . 2C. Differentials; Approximations dx dy dz + + x y z 2y dx − 2x dy c) dz = (x + y)2 b) dw = 3x2 y 2 z dx + 2x3 yz dy + x3 y 2 dz 2C-1 a) dw = t du − u dt d) dw = √ t t2 − u 2 2C-2 The volume is V = xyz; so dV = yz dx + xz dy + xy dz. For x = 5, y = 10, z = 20, ΔV ≈ dV = 200 dx + 100 dy + 50 dz, from which we see that |ΔV | ≤ 350(.1); therefore V = 1000 ± 35. 2C-3 a) A = 12 ab sin θ. Therefore, dA = 12 (b sin θ da + a sin θ db + ab cos θ dθ). √ √ b) dA = 21 (2 · 12 da + 1 · 21 db + 1 · 2 · 12 3 dθ) = 12 (da + 12 db + 3 dθ); therefore most sensitive to θ, least senstitive to b, since dθ and db have respectively the largest and smallest coefficients. c) dA = 21 (.02 + .01 + 1.73(.02) ≈ 12 (.065) ≈ .03 k kT kT ; therefore dP = dT − 2 dV V V V k dT − P dV b) V dP + P dV = k dT ; therefore dP = . V c) Substituting P = kT /V into (b) turns it into (a). � � dt du dv dt du dv dw therefore dw = w2 2 + 2 + 2 . 2C-5 a) − 2 = − 2 − 2 − 2 ; w t u v t u v u du + 2v dv b) 2u du + 4v dv + 6w dw = 0; therefore dw = − . 3w 2C-4 a) P = 2D. Gradient; Directional Derivative √ � i−j df �� 3 2 = (3 i + 6 j ) · √ 2D-1 a) ∇f = 3x i + 6y j ; (∇f )P = 3 i + 6 j ; =− ds �u 2 2 � x xy dw �� 1 y i + 2j − 2k =− b) ∇w = i + j − 2 k ; (∇w)P = − i +2 j +2 k ; = (∇w)P · z z z ds �u 3 3 2 2 c) ∇z = (sin y − y sin x) i + (x cos � y + cos x) j ; (∇z)P = i + j ; dz �� 1 −3 i + 4 j = = (i + j) · � ds 5 5 u 2. d) ∇w = 2i + 3j ; 2t + 3u PARTIAL DIFFERENTIATION 3 � 4i − 3 j dw �� 1 = (2 i + 3 j ) · =− � ds u 5 5 (∇w)P = 2 i + 3 j ; e) ∇f = 2(u + 2v + 3w)(� i + 2 j + 3 k ); (∇f )P = 4( i + 2 j + 3 k ) df �� −2 i + 2 j − k 4 = 4( i + 2 j + 3 k ) · =− � ds u 3 3 4i − 3j ; (∇w)P = 4 i − 3 j 2D-2 a) ∇w = 4x − 3y � 4i − 3j dw �� = (4 i − 3 j ) · u has maximum 5, in the direction u = , ds �u 5 and minimum −5 in the opposite direction. � 3i + 4j dw �� = 0 in the directions ± . ds �u 5 b) ∇w = (∇w)P = �1, 3�, 0�; � �y + z, x + z, x + y�; √ √ dw �� dw �� i + 3j i + 3j √ ; min ; max 10, direction 10, direction − √ = = − � ds ��u ds 10 10 u dw �� −3 i + j + c k (for all c) = 0 in the directions u = ± √ � ds u 10 + c2 c) ∇z =� 2 sin(t − u) cos(t − u)( i − j ) = sin 2(�t − u)( i − j ); (∇z)P = i − j ; √ √ dz �� i−j −i + j dz �� = 2, direction √ ; = − 2, direction − √ max min ; ds ��u ds �u 2 2 i+j dz �� = 0 in the directions ± √ ds �u 2 2D-3 a) ∇f = �y 2 z 3 , 2xyz 3 , 3xy 2 z 2 �; (∇f )P = �4, 12, 36�; tangent plane at P : x + 3y + 9z = 18 b) ∇f = �2x, 8y, 18z�; normal at P : �1, 4, 9�, normal at P : �1, 3, 9�; tangent plane: x + 4y + 9z = 14. c) (∇w)P = �2x0 , 2y0 , −2z0 �; tangent plane: x0 (x − x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0, or x0 x + y0 y − z0 z = 0, since x20 + y02 − z02 = 0. 2D-4 a) ∇T = 2x i + 2y j ; x2 + y 2 (∇T )P = 2i + 4j ; 5 T is increasing at P most rapidly in the direction of (∇T )P , which is i + 2j √ . 5 2 i + 2j b) |∇T | = √ = rate of increase in direction √ . Call the distance to go Δs, then 5 5√ √ .2 5 2 5 √ Δs = .20 ⇒ Δs = = ≈ .22. 2 10 5 � 6 dT �� 2i + 4j i + j · √ = √ ; c) = (∇T )P · u = ds �u 5 2 5√ 2 √ 6 5 2 √ Δs = .12 ⇒ Δs = (.12) ≈ (.10)( 2) ≈ .14 6 5 2 2i − j d) In the directions orthogonal to the gradient: ± √ 5 4 S. 18.02 SOLUTIONS TO EXERCISES 2D-5 a) isotherms = the level surfaces x2 + 2y 2 + 2z 2 = c, which are ellipsoids. b) ∇T = �2x, 4y, 4z�; (∇T )P = �2, 4, 4�; |(∇T )P | = 6; for most rapid decrease, use direction of −(∇T )P : − 13 �1, 2, 2� c) let Δs be distance to go; then −6(Δs) = −1.2; � �1, −2, 2� dT �� 2 = (∇T )P · u = �2, 4, 4� · d) = ; ds �u 3 3 Δs ≈ .2 2 Δs ≈ .10 ⇒ Δs ≈ .15. 3 2D-6 ∇uv = �(uv)x , (uv)y � = �uvx +vux , uvy +vuy � = �uvx , uvy �+�vux +vuy � = u∇v+v∇u � � � d(uv) �� dv �� du �� ∇(uv) = u∇v +v∇u ⇒ ∇(uv)·u = u∇v ·u+v∇u·u ⇒ = u +v . ds �u ds �u ds �u 2D-7 At P , let ∇w = a i + b j . Then √ i+j =2 ⇒ a+b=2 2 ai + bj · √ 2 √ i−j =1 ⇒ a−b= 2 ai + bj · √ 2 √ √ Adding and subtracting the equations on the right, we get a = 23 2, b = 12 2. 2D-8 We have P (0, 0, 0) = 32; we wish to decrease it to 31.1 by traveling the shortest distance from the origin 0; for this we should travel in the direction of −(∇P )0 . ∇P = �(y + 2)ez , (x + 1)ez , (x + 1)(y + 2)ez �; (∇P )0 = �2, 1, 2�. |(∇P )0 | = 3. Since (−3) · (Δs) = −.9 ⇒ Δs = .3, we should travel a distance .3 in the direction 1 of −(∇P )0 . Since | − �2, 1, 2�| = 3, the distance .3 will be 10 of the distance from (0, 0, 0) to (−2, −1, −2), which will bring us to (−.2, −.1, −.2). � Δw dw �� ≈ 2D-9 In these, we use : we travel in the direction u from a given point P to ds �u Δs the nearest level curve C; then Δs is the distance traveled (estimate it by using the unit distance), and Δw is the corresponding change in w (estimate it by using the labels on the level curves). a) The direction of ∇f is perpendicular to the level curve at A, in the increasing sense (the “uphill” direction). The magnitude of ∇f is the directional derivative in that direction: 1 Δw ≈ = 2. from the picture, Δs .5 � � ∂w dw �� ∂w dw �� b), c) , , so B will be where i is tangent to the level curve = = ∂x ds � i ∂y ds � j and C where j is tangent to the level curve. � � ∂w dw �� −1 ∂w dw �� −1 Δw Δw d) At P, = ≈ = −.6; = ≈ = −1. ≈ ≈ � � P ∂x ds i Δs 5/3 ∂y ds j Δs 1 B � 1 Δw dw �� 1 2 3 4 ≈ e) If u is the direction of i + j , we have ≈ =2 5 C ds� �u Δs .5 Q Δw dw �� −1 B ≈ f) If u is the direction of i − j , we have ≈ = −.8 � ds u Δs 5/4 g) The gradient is 0 at a local extremum point: here at the point 1 marked giving the location of the hilltop. C A 2. PARTIAL DIFFERENTIATION 2E. Chain Rule 2E-1 ∂w dx ∂w dy ∂w dz dw = + + = yz · 1 + xz · 2t + xy · 3t2 = t5 + 2t5 + 3t5 = 6t5 dt ∂x dt ∂y dt ∂z dt dw (ii) w = xyz = t6 ; = 6t5 dt a) (i) ∂w dx ∂w dy dw = + = 2x(− sin t) − 2y(cos t) = −4 sin t cos t dt ∂x dt ∂y dt dw (ii) w = x2 − y 2 = cos2 t − sin2 t = cos 2t; = −2 sin 2t dt b) (i) 2u 2v dw = 2 (−2 sin t) + 2 (2 cos t) = − cos t sin t + sin t cos t = 0 2 dt u +v u + v2 dw (ii) w = ln(u2 + v 2 ) = ln(4 cos2 t + 4 sin2 t) = ln 4; = 0. dt c) (i) 2E-2 a) The value to� the �point (x(0), y(0)) = (1, � t = 0 corresponds � � � 0) = P . dw �� ∂w �� dx �� ∂w �� dy �� = + = −2 sin t + 3 cos t = 3. dt �0 ∂x �P dt �0 ∂y �P dt �0 0 b) ∂w dx ∂w dy dw = + = y(− sin t) + x(cos t) = − sin2 t + cos2 t = cos 2t. dt ∂x dt ∂y dt π π nπ dw = 0 when 2t = + nπ, therefore when t = + . dt 2 4 2 c) t = 1 corresponds � to the �point (x(1), � y(1), z(1)) � = (1, 1, 1). df �� dx �� dy �� dz �� =1· −1· + 2 · � = 1 · 1 − 1 · 2 + 2 · 3 = 5. dt � dt � dt � dt 1 d) 1 1 1 dx dy dz df = 3x2 y + (x3 + z) +y = 3t4 · 1 + 2x3 · 2t + t2 · 3t2 = 10t4 . dt dt dt dt 2E-3 a) Let w = uv, where u = u(t), v = v(t); b) du dv dw d(uvw) = vw + uw + uv ; dt dt dt dt dw ∂w du ∂w dv du dv = + =v +u . dt ∂u dt ∂v dt dt dt e2t sin t + 2te2t sin t + te2t cos t 2E-4 The values u = 1, v = 1 correspond to the point x = 0, y = 1. At this point, ∂w ∂w ∂x ∂w ∂y = + = 2 · 2u + 3 · v = 2 · 2 + 3 = 7. ∂u ∂x ∂u ∂y ∂u ∂w ∂x ∂w ∂y ∂w = + = 2 · (−2v) + 3 · u = 2 · (−2) + 3 · 1 = −1. ∂v ∂x ∂v ∂y ∂v a) wr = wx xr + wy yr = wx cos θ + wy sin θ wθ = wx xθ + wy yθ = wx (−r sin θ) + wy (r cos θ) Therefore, (wr )2 + (wθ /r)2 = (wx )2 (cos2 θ + sin2 θ) + (wy )2 (sin2 θ + cos2 θ) + 2wx wy cos θ sin θ − 2wx wy sin θ cos θ = (wx )2 + (wy )2 . 2E-5 5 6 S. 18.02 SOLUTIONS TO EXERCISES √ b) The point r = 2, θ = π/4 in polar coordinates corresponds in rectangular coordi­ nates to the point x = 1, y = 1. Using the chain rule equations in part (a), wr = wx cos θ + wy sin θ; wθ = wx (−r sin θ) + wy (r cos θ) but evaluating all the partial derivatives at the point, we get √ √ √ √ √ √ wθ wr = 2 · 12 2 − 1 · 21 2 = 21 2; = 2(− 12 ) 2 − 12 2 = − 32 2; r 1 1 9 2 2 (wr ) + (wθ ) = 2 + 2 = 5; (wx )2 + (wy )2 = 22 + (−1)2 = 5. r 2E-6 wu = wx · 2u + wy · 2v; wv = wx · (−2v) + wy · 2u, by the chain rule. Therefore (wu )2 + (wv )2 = [4u2 (wx ) + 4v 2 (wy )2 + 4uvwx wy ] + [4v 2 (wx ) + 4u2 (wy )2 − 4uvwx wy ] = 4(u2 + v 2 )[(wx )2 + (wy )2 ]. 2E-7 By the chain rule, fu = fx xu + fy yu , fv �= fx xv +�fy yv ; xu xv �fu fv � = �fx fy � yu yv therefore 2E-8 a) By the chain rule for functions of one variable, ∂u y ∂w ∂u 1 ∂w = f ′ (u) · = f ′ (u) · − 2 ; = f ′ (u) · = f ′ (u) · ; ∂x ∂x x ∂y ∂y x Therefore, ∂w ∂w y y +y = f ′ (u) · − + f ′ (u) · = 0. x ∂x ∂y x x 2F. Maximum-minimum Problems 2F-1 In these, denote by D = x2 + y 2 + z 2 the square of the distance from the point (x, y, z) to the origin; then the point which minimizes D will also minimize the actual distance. 1 1 , we get on substituting, D = x2 + y 2 + . with x and y xy xy independent; setting the partial derivatives equal to zero, we get 1 1 1 1 , 2y 2 = . Dx = 2x − 2 = 0; Dy = 2y − 2 = 0; or 2x2 = x y y x xy xy 1 Solving, we see first that x2 = = y 2 , from which y = ±x. 2xy a) Since z 2 = If y = x, then x4 = 12 and x = y = 2−1/4 , and so z = 21/4 ; if y = −x, then x4 = − 21 and there are no solutions. Thus the unique point is (1/21/4 , 1/21/4 , 21/4 ). b) Using the relation x2 = 1 + yz to eliminate x, we have D = 1 + yz + y 2 + z 2 , with y and z independent; setting the partial derivatives equal to zero, we get Dy = 2y + z = 0, Dz = 2z + y = 0; solving, these equations only have the solution y = z = 0; therefore x = ±1, and there are two points: (±1, 0, 0), both at distance 1 from the origin. 2F-2 Letting x be the length of the ends, y the length of the sides, and z the height, we have total area of cardboard A = 3xy + 4xz + 2yz, volume V = xyz = 1. Eliminating z to make the remaining variables independent, and equating the partials to zero, we get 2. PARTIAL DIFFERENTIATION 2 4 + ; Ax = 3y − y x From these last two equations, we get 2 4 3xy = , 3xy = ⇒ x y A = 3xy + ⇒ 3x3 = 1 ⇒ x= 2 = 0, x2 2 4 = x y 7 Ay = 3x − ⇒ 4 = 0. y2 y = 2x 2 1 32/3 3 , y = , z = = = ; 1/3 1/3 xy 2 3 3 2 · 31/3 1 therefore the proportions of the most economical box are x : y : z = 1 : 2 : 32 . 2F-5 The cost is C = xy + xz + 4yz + 4xz, where the successive terms represent in turn the bottom, back, two sides, and front; i.e., the problem is: minimize: C = xy + 5xz + 4yz, with the constraint: xyz = V = 2.5 Substituting z = V /xy into C, we get C = xy + 5V 4V + ; y x ∂C 4V =y− 2, ∂x x ∂C 5V =x− 2 . ∂y y We set the two partial derivatives equal to zero and solving the resulting equations simulta­ 16V neously, by eliminating y; we get x3 = = 8, (using V = 5/2), so x = 2, y = 52 , z = 12 . 5 2G. Least-squares Interpolation 2G-1 Find y = mx + b that best fits (1, 1), (2, 3), (3, 2) . D = (m + b − 1)2 + (2m + b − 3)2 + (3m + b − 2)2 ∂D = 2(m + b − 1) + 4(2m + b − 3) + 6(3m + b − 2) = 2(14m + 6b − 13) ∂m ∂D = 2(m + b − 1) + 2(2m + b − 3) + 2(3m + b − 2) = 2(6m + 3b − 6). ∂b � 14m + 6b = 13 ∂D ∂D Thus the equations = 0 and = 0 are , whose solution is 6m + 3b = 6 ∂m ∂b m = 12 , b = 1, and the line is y = 12 x + 1 . � 2 2G-4 D = i (a �+ bxi + cyi − zi ) . The equations are ∂D/∂a = � 2(a + bxi + cyi − zi ) = 0 ∂D/∂b = � 2xi (a + bxi + cyi − zi ) = 0 ∂D/∂c = 2yi (a + bxi + cyi − zi ) = 0 Cancel the 2’s; the equations � � na + ( xi )b + ( yi )c � � � ( xi )a + ( x2i )b + ( xi yi )c � � � ( yi )a + ( xi yi )b + ( yi2 )c become (on the right, x = [x1 , . . . , xn ], 1 = [1, . . . , 1], etc.) � = zi n a + (x · 1) b + (y · 1) c = z · 1 � = x i zi or (x · 1) a + (x · x) b + (x · y) c = x · z � = y i zi (y · 1) a + (x · y )b + (y · y) c = y · z 8 S. 18.02 SOLUTIONS TO EXERCISES 2H. Max-min: 2nd Derivative Criterion; Boundary Curves 2H-1 a) fx = 0 : 2x − y = 3; fy = 0 : −x − 4y = 3 A = fxx = 2; B = fxy = −1; C = fyy = −4; b) fx = 0 : 6x + y = 1; fy = 0 : x + 2y = 2 A = fxx = 6; B = fxy = 1; C = fyy = 2; c) fx = 0 : 8x3 − y = 0; critical point: (1, −1) AC − B 2 = −9 < 0; saddle point critical point: (0, 1) AC − B 2 = 11 > 0; local minimum fy = 0 : 2y − x = 0; eliminating y, we get 16x − x = 0, or x(16x − 1) = 0 ⇒ x = 0, x = 14 , x = − 41 , giving the critical points (0, 0), ( 14 , 18 ), (− 14 , − 18 ). 3 2 Since fxx = 24x2 , fxy = −1, (0, 0) : Δ = −1 (saddle); fyy = 2, we get for the three points respectively: ( 14 , 18 ) : Δ = 2 (minimum); d) fx = 0 : 3x2 − 3y = 0; (− 14 , − 18 ) : Δ = 2 (minimum) fy = 0 : −3x + 3y 2 = 0. Eliminating y gives −x + x4 = 0, or x(x3 − 1) = 0 ⇒ x = 0, y = 0 or x = 1, y = 1. Since fxx = 6x, fxy = −3, fyy = 6y, we get for the two critical points respectively: (0, 0) : AC − B 2 = −9 (saddle); (1, 1) : AC − B 2 = 27 (minimum) e) fx = 0 : 3x2 (y 3 + 1) = 0; fy = 0 : 3y 2 (x3 + 1) = 0; solving simultaneously, we get from the first equation that either x = 0 or y = −1; finding in each case the other coordinate then leads to the two critical points (0, 0) and (−1, −1). Since fxx = 6x(y 3 + 1), fxy = 3x2 · 3y 2 , (−1, −1) : AC − B 2 = −9 (saddle); fyy = 6y(x3 + 1) , we have (0, 0) : AC − B 2 = 0, test fails. (By studying the behavior of f (x, y) on the lines y = mx, for different values of m, it is possible to see that also (0, 0) is a saddle point.) 2H-3 The region R has no critical points; namely, the equations fx = 0 and fy = 0 are 2x + 2 = 0, 2y + 4 = 0 ⇒ x = −1, y = −2, but this point is not in R. We therefore investigate the diagonal boundary of R, using the parametrization x = t, y = −t. Restricted to this line, f (x, y) becomes a function of t alone, which we denote by g(t), and we look for its maxima and minima. g(t) = f (t, −t) = 2t2 − 4t − 1; g ′ (t) = 4t − 2, which is 0 at t = 1/2. This point is evidently a minimum for g(t); there is no maximum: g(t) tends to ∞. Therefore for f (x, y) on R, the minimum occurs at the point (1/2, −1/2), and there is no maximum; f (x, y) tends to infinity in different directions in R. 2. 2H-4 PARTIAL DIFFERENTIATION 9 We have fx = y − 1, fy = x − 1, so the only critical point is at (1, 1). a) On the two sides of the boundary, the function f (x, y) becomes respectively y = 0 : f (x, y) = −x + 2; x = 0 : f (x, y) = −y + 2. Since the function is linear and decreasing on both sides, it has no minimum points (infor­ mally, the minimum is −∞). Since f (1, 1) = 1 and f (x, x) = x2 − 2x + 2 → ∞ as x → ∞, the maximum of f on the first quadrant is ∞. b) Continuing the reasoning of (a) to find the maximum and minimum points of f (x, y) on the boundary, on the other two sides of the boundary square, the function f (x, y) becomes y = 2 : f (x, y) = x x = 2 : f (x, y) = y Since f (x, y) is thus increasing or decreasing on each of the four sides, the maximum and minimum points on the boundary square R can only occur at the four corner points; evaluating f (x, y) at these four points, we find f (0, 0) = 2; f (2, 2) = 2; f (2, 0) = 0; As in (a), since f (1, 1) = 1, maximum points of f on R: (0, 0) and (2, 2); x 2 2 -y y 2 2 -x f (0, 2) = 0. minimum points: (2, 0) and (0, 2). c) The data indicates that (1, 1) is probably a saddle point. Confirming this, we have fxx = 0, fxy = 1, fyy = 0 for all x and y; therefore AC − B 2 = −1 < 0, so (1, 1) is a saddle point, by the 2nd-derivative criterion. 2H-5 Since f (x,√ y) is linear, it will not have critical points: namely, for all x and y we have fx = 1, fy = 3. So any maxima or minima must occur on the boundary circle. We parametrize the circle by x = cos θ, y = sin θ; restricted to this boundary circle, f (x, y) becomes a function of θ alone which we call g(θ): √ g(θ) = f (cos θ, sin θ) = cos θ + 3 sin θ + 2. Proceeding in the usual way to find the maxima and minima of g(θ), we get √ √ g ′ (θ) = − sin θ + 3 cos θ = 0, or tan θ = 3. π 4π and ; evaluating g at these two 3 3 points, we get g(π/3) = 4 (the maximum), and g(4π/3) = 0 (the √ minimum). 3/2), while the minimum Thus the maximum of f (x, y) in the circular disc R is at (1/2, √ is at (−1/2, − 3/2). It follows that the two critical points of g(θ) are θ = 2H-6 a) Since z = 4 − x − y, the problem is to find on R the maximum and minimum of the total area f (x, y) = xy + 14 (4 − x − y)2 where R is the triangle given byR : 0 ≤ x, 0 ≤ y, x + y ≤ 4. 0 4 x y x y To find the critical points of f (x, y), the equations fx = 0 and fy = 0 are respectively y − 21 (4 − x − y) = 0; z x − 12 (4 − x − y) = 0, which imply first that x = y, and from this, x− 21 (4−2x); the unique solution is x = 1, y = 1. z 2 10 S. 18.02 SOLUTIONS TO EXERCISES The region R is a triangle, on whose sides f (x, y) takes respectively the values bottom: y = 0; f = 14 (4 − x)2 ; diagonal y = 4 − x; f = x(4 − x). left side: x = 0; f = 41 (4 − y)2 ; On the bottom and side, f is decreasing; on the diagonal, f has a maximum at x = 2, y = 2. Therefore we need to examine the three corner points and (2, 2) as candidates for maximum and minimum points, as well as the critical point (1, 1). We find f (0, 0) = 4; f (4, 0) = 0; f (0, 4) = 0; f (2, 2) = 4 4 (4 y) x (4 x ) 2 4 x )2 (4 4 4 f (1, 1) = 2. It follows that the critical point is just a saddle point; to get the maximum total area 4, make x = y = 0, z = 4, or x = y = 2, z = 0, either of which gives a point “rectangle” and a square of side 2; for the minimum total area 0, take for example x = 0, y = 4, z = 0, which gives a “rectangle” of length 4 with zero area, and a point square. b) We have fxx = 12 , fxy = 32 , fyy = 12 for all x and y; therefore AC − B 2 = −2 < 0, so 2 x 2- 6x+ 4 (1, 1) is a saddle point, by the 2nd-derivative criterion. 2 2H-7 a) fx = 4x − 2y − 2, fy = −2x + 2y; setting these = 0 and solving simultaneously, we get x = 1, y = 1, which is therefore the only critical point. On the four sides of the boundary rectangle R, the function f (x, y) becomes: on y = −1 : f (x, y) = 2x2 + 1; on y = 2 : f (x, y) = 2x2 − 6x + 4 on x = 0 : f (x, y) = y 2 ; on x = 2 : f (x, y) = y 2 − 4y + 4 y 2 4 y +4 y2 2 1 2 x 2+ 1 By one-variable calculus, f (x, y) is increasing on the bottom and decresing on the right side; on the left side it has a minimum at (0, 0), and on the top a minimum at ( 32 , 2). Thus the maximum and minimum points on the boundary rectangle R can only occur at the four corner points, or at (0, 0) or ( 23 , 2). At these we find: f (0, −1) = 1; f (0, 2) = 4; f (2, −1) = 9; f (2, 2) = 0; f ( 32 , 2) = − 21 , f (0, 0) = 0. At the critical point f (1, 1) = −1; comparing with the above, it is a minimum; therefore, maximum point of f (x, y) on R: (2, −1) minimum point of f (x, y) on R: (1, 1) b) We have fxx = 4, fxy = −2, fyy = 2 for all x and y; therefore AC − B 2 = 4 > 0 and A = 4 > 0, so (1, 1) is a minimum point, by the 2nd-derivative criterion. 2I. Lagrange Multipliers 2I-1 Letting P : (x, y, z) be the point, in both problems we want to maximize V = xyz, subject to a constraint f (x, y, z) = c. The Lagrange equations for this, in vector form, are ∇(xyz) = λ · ∇f (x, y, z), f (x, y, z) = c. a) Here f = c is x + 2y + 3z = 18; equating components, the Lagrange equations become yz = λ, xz = 2λ, xy = 3λ; x + 2y + 3z = 18. To solve these symmetrically, multiply the left sides respectively by x, y, and z to make them equal; this gives λx = 2λy = 3λz, or x = 2y = 3z = 6, since the sum is 18. 2. PARTIAL DIFFERENTIATION 11 We get therefore as the answer x = 6, y = 3, z = 2. This is a maximum point, since if P lies on the triangular boundary of the region in the first octant over which it varies, the volume of the box is zero. b) Here f = c is x2 + 2y 2 + 4z 2 = 12; equating components, the Lagrange equations become yz = λ · 2x, xz = λ · 4y, xy = λ · 8z; x2 + 2y 2 + 4z 2 = 12. To solve these symmetrically, multiply the left sides respectively by x, y, and z to make them equal; this gives λ · 2x2 = λ · 4y 2 = λ · 8z 2 , x2 = 2y 2 = 4z 2 = 4, since the sum is 12. √ We get therefore as the answer x = 2, y = 2, z = 1. This is a maximum point, since if P lies on the boundary of the region in the first octant over which it varies (1/8 of the ellipsoid), the volume of the box is zero. √ 2I-2 Since we want to minimize x2 + y 2 + z 2 , subject to the constraint x3 y 2 z = 6 3, the Lagrange multiplier equations are √ 2x = λ · 3x2 y 2 z, 2y = λ · 2x3 yz, 2z = λ · x3 y 2 ; x3 y 2 z = 6 3. or To solve them symmetrically, multiply the first three equations respectively by x, y, and z, then divide them through respectively by 3, 2, and 1; this makes the right sides equal, so that, after canceling 2 from every numerator, we get √ √ y2 x2 = = z 2 ; therefore x = z 3, y = z 2. 3 2 √ √ 3 √ 3 2 2 Substituting into √ x y z = 6 3, we get 3 3z · 2z · z = 6 3, which gives as the answer, √ x = 3, y = 2, z = 1. This is clearly a minimum, since if P is near one of the coordinate planes, one of √ the variables is close to zero and therefore one of the others must be large, since x3 y 2 z = 6 3; thus P will be far from the origin. 2I-3 Referring to the solution of 2F-2, we let x be the length of the ends, y the length of the sides, and z the height, and get total area of cardboard A = 3xy + 4xz + 2yz, volume V = xyz = 1. The Lagrange multiplier equations ∇A = λ · ∇(xyz); xyz = 1, then become 3y + 4z = λyz, 3x + 2z = λxz, 4x + 2y = λxy, xyz = 1. To solve these equations for x, y, z, λ, treat them symmetrically. Divide the first equation through by yz, and treat the next two equations analogously, to get 3/z + 4/y = λ, 3/z + 2/x = λ, 4/y + 2/x = λ, which by subtracting the equations in pairs leads to 3/z = 4/y = 2/x; setting these all equal to k, we get x = 2/k, y = 4/k, z = 3/k, which shows the proportions using least cardboard are x : y : z = 2 : 4 : 3. To find the actual values of x, y, and z, we set 1/k = m; then substituting into xyz = 1 gives (2m)(4m)(3m) = 1, from which m3 = 1/24, m = 1/2 · 31/3 , giving finally 1 2 3 x = 1/3 , y = 1/3 , z = . 3 3 2 · 31/3 12 S. 18.02 SOLUTIONS TO EXERCISES 2I-4 The equations for the cost C and the volume V are xy +4yz +6xz = C and xyz = V . The Lagrange multiplier equations for the two problems are a) yz = λ(y + 6z), b) xz = λ(x + 4z), y + 6z = µ · yz, xy = λ(4y + 6x); x + 4z = µ · xz, xy + 4yz + 6xz = 72 4y + 6x = µ · xy; xyz = 24 The first three equations are the same in both cases, since we can set µ = 1/λ. Solving the first three equations in (a) symmetrically, we multiply the equations through by x, y, and z respectively, which makes the left sides equal; since the right sides are therefore equal, we get after canceling the λ, xy + 6xz = xy + 4yz = 4yz + 6xz, which implies xy = 4yz = 6xz. a) Since the sum of the three equal products is 72, by hypothesis, we get xy = 24, yz = 6, xz = 4; from the first two we get x = 4z, and from the first and third we get y = 6z, which lead to the solution x = 4, y = 6, z = 1. b) Dividing xy = 4yz = 6xz by xyz leads after cross-multiplication to x = 4z, y = 6z; since by hypothesis, xyz = 24, again this leads to the solution x = 4, y = 6, z = 1. 2J. Non-independent Variables � ∂w ∂y � means that x is the dependent variable; get rid of it by writing � � ∂w w = (z − y)2 + y 2 + z 2 = z + z 2 . This shows that = 0. ∂y z � � ∂w b) To calculate , once again x is the dependent variable; as in part (a), we ∂z y � � ∂w have w = z + z 2 and so = 1 + 2z. ∂z y 2J-1 a) z � � � ∂x y ∂x + 2y; so =− ; 2J-2 a) Differentiating z = x + y w.r.t. y: 0 = 2x ∂y z x � � � � ∂y � z � ∂x −y ∂w By the chain rule, = 2x + 2y = 2x + 2y = 0. ∂y z ∂y z x � � � � ∂x ∂x 1 Differentiating z = x2 + y 2 with respect to z: 1 = 2x ; so = ; ∂z y ∂z y 2x � � � � ∂w ∂x By the chain rule, = 2x + 2z = 1 + 2z. ∂z y ∂z y 2 2 � b) Using differentials, dw = 2xdx + 2ydy + 2zdz, dz = 2xdx + 2ydy; since the independent variables are y and z, we eliminate dx by substracting the second equation from the first, which�gives� dw =�0 dy + (1 + 2z) dz; � ∂w ∂w = 0, = 1 + 2z. therefore by D2, we get ∂y z ∂z y 2. � ∂w ∂t � PARTIAL DIFFERENTIATION 13 , we see that y is the dependent variable; solving for it, we � � � � zt ∂y z ∂w 3 get y = ; using the chain rule, = x − z 2 = x3 − z 2 = x2 z − z 2 . x ∂t x,z ∂t x,z x � � ∂w xy b) Similarly, means that t is the dependent variable; since t = , we ∂z x,y z � � � � ∂t −xy ∂w = −2zt − z 2 = −2zt − z 2 · 2 = −zt. have by the chain rule, ∂z x,y ∂z x,y z 2J-3 a) To calculate x,z 2J-4 The differentials are calculated in equation (4). a) Since x, z, t are independent, we eliminate dy by solving the second equation for x dy, substituting this into the first equation, and grouping terms: � � ∂w dw = 2x2 y dx + (x2 z − z 2 )dt + (x2 t − 2zt)dz, which shows by D2 that = x2 z − z 2 . ∂t x,z b) Since x, y, z are independent, we eliminate dt by solving the second equation for z dt, substituting this into the first equation, and grouping terms: � � ∂w 2 3 dw = (3x y − zy)dx + (x − zx)dy − zt dz, which shows by D2 that = −zt. ∂z x,y 2J-5 a) If pv = nRT , then � b) Similarly, we have � 2J-6 a) � � ∂S ∂p � ∂S ∂T � v v = Sp + ST · � = ST + Sp · � � ∂T ∂p � ∂p ∂T � v v = Sp + ST · v . nR = ST + Sp · nR . v � ∂w ∂v = 3u2 − v 2 − u · 2v = 3u2 − v 2 − 2uv. ∂u x ∂u x � � � � ∂v ∂w = −u · 2v = −2uv. ∂x u ∂x u b) dw = (3u2 − v 2 )du − 2uvdv; du = x dy + y dx; dv = du + dx; for both derivatives, u and x are the independent variables, so we eliminate dv, getting 2 2 dw = (3u2 − v 2 )du − � 2uv(du � + dx)�= (3u � − v − 2uv)du − 2uv dx, ∂w ∂w whose coefficients by D2 are and . ∂u x ∂x u 2J-7 Since we need both derivatives for the gradient, we use differentials. df = 2dx + dy − 3dz at P ; dz = 2x dx + dy = 2 dx + dy at P ; the independent variables are to be x and z, so we eliminate dy, getting df = 0 dx − 2 dz at the point (x, z) = (1, 1). So ∇g = �0, −2� at (1, 1). � � ∂w 2J-8 To calculate , note that r and θ are independent. Therefore, ∂r �θ � � � � � ∂w ∂w ∂w ∂x ∂x = + · . Now, x = r cos θ, so = cos θ . ∂r ∂r ∂x ∂r ∂r θ θ � �θ ∂w r −x r − x cos θ = √ + √ · cos θ = √ 2 2 2 2 ∂r θ r −x r −x r 2 − x2 Therefore 14 S. 18.02 SOLUTIONS TO EXERCISES = r sin2 θ r − r cos2 θ = = | sin θ|. r| sin θ| r| sin θ| 2K. Partial Differential Equations 1 2 ln(x2 + y 2 ). If (x, y) �= (0, 0), then � � ∂ x y 2 − x2 ∂ = , wxx = (wx ) = ∂x ∂x � x2 + y 2 � (x2 + y 2 )2 ∂ ∂ y x2 − y 2 wyy = (wy ) = = , 2 2 ∂y ∂y x + y (x2 + y 2 )2 Therefore w satisfies the two-dimensional Laplace equation, wxx + wyy = 0; we exclude the point (0, 0) since ln 0 is not defined. 2K-1 w = � ∂ � ∂ (wx ) = 2x · n(x2 + y 2 + z 2 )n−1 ∂x ∂x = 2n(x2 + y 2 + z 2 )n−1 + 4x2 n(n − 1)(x2 + y 2 + z 2 )n−2 2K-2 If w = (x2 + y 2 + z 2 )n , then We get wyy and wzz by symmetry; adding and combining, we get wxx + wyy + wzz = 6n(x2 + y 2 + z 2 )n−1 + 4(x2 + y 2 + z 2 )n(n − 1)(x2 + y 2 + z 2 )n−2 = 2n(2n + 1)(x2 + y 2 + z 2 )n−1 , which is identically zero if n = 0, or if n = −1/2. 2K-3 a) w = ax2 + bxy + cy 2 ; wxx = 2a, wxx + wyy = 0 ⇒ wyy = 2c. 2a + 2c = 0, or c = −a. Therefore all quadratic polynomials satisfying the Laplace equation are of the form ax2 + bxy − ay 2 = a(x2 − y 2 ) + bxy; i.e., linear combinations of the two polynomials f (x, y) = x2 − y 2 and g(x, y) = xy . 1 wtt . So c2 w = f (x + ct) + g(x − ct) ⇒ wxx = f ′′ (x + ct) + g ′′ (x − ct) ⇒ wt = cf ′ (x + ct) + −cg ′ (x − ct). ⇒ wtt = c2 f ′′ (x + ct) + c2 g ′′ (x − ct) = c2 wxx , which shows w satisfies the wave equation. 2K-4 The one-dimensional wave equation is wxx = 1 wt . So if w(x, t) = sin kxer t, then α2 = ert · k 2 (− sin kx) = −k 2 w. wt = rert sin kx = r w. 2K-5 The one-dimensional heat equation is wxx = wxx 1 r w, or r = −α2 k 2 . α2 However, from the additional condition that w = 0 at x = 1, we must have sin k ert = 0 ; Therefore, we must have −k 2 w = Therefore sin k = 0, and so k = nπ, where n is an integer. To see what happens to w as t → ∞, we note that since | sin kx| ≤ 1, |w| = ert | sin kx| ≤ ert . Now, if k = � 0, then r = −α2 k 2 is negative and ert → 0 as t → ∞; therefore |w| → 0. 2. PARTIAL DIFFERENTIATION 15 Thus w will be a solution satisfying the given side conditions if k = nπ, where n is a non-zero integer, and r = −α2 k 2 .