\f\f3. a) potential function = x2 +y 2 +z 2 ; magnetic force M = y 2 zi+z 2 xj+x2 yk i) Grad = i+ j+ k x y z Note that : = 2x = 2y = 2z x y z Hence, Grad = 2xi + 2yj + 2zk M M M ii) Div M = + + x y z M M M Note that : =0 =0 =0 x y z Hence, Div M = 0 + 0 + 0 = 0. iii) Div Grad = Div (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) (2x) (2y) (2z) + + x y z =2+2+2 = = 6. Hence Div Grad = 6. iv) Curl Grad = Curl (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) i = /x 2x = j /y 2y k /z 2z (2z) (2y) y z (2x) (2z) z x i+ j+ (2y) (2x) z y k = 0i + 0j + 0k = 0 Hence, Curl Grad = 0. v) Curl M = Grad M = Grad (y 2 zi + z 2 xj + x2 yk) i = /x y2 z = j /y z2x k /z x2 y (x2 y) (z 2 x) y z i+ (y 2 z) (x2 y) z x j+ (z 2 x) (y 2 z) z y = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k Hence, curl M = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k. 1 k vi) Div Curl M = Div (y 2 zi + z 2 xj + x2 yk) = Grad (y 2 zi + z 2 xj + x2 yk) (x2 2zx) (y 2 2xy) (z 2 2yz) + + x y z = (2x 2z) + (2y 2x) + (2z 2y) = = (2x 2x) + (2y 2y) + (2z 2z) = 0 b) Hence, Div Curl M = 0. i) Curl Grad f f f f i+ j+ k x y z f f f = Grad i+ j+ k x y z = Curl i /x = f x = j /y k /z f y f z f ( f ) ( y ) z y x ( f ) ( f ) x z z x i+ j+ ( f ) y x ( f ) x y k = ( 2 f /zy 2 f /yz)i + ( 2 f /xz 2 f /zz)j + ( 2 f /yx 2 f /xy)k This is not the zero vector for any f . However, Curl Grad f = 0, for example, for any open ball in R3 , in which f is continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 f /zy = 2 f /yz, and 2 f /xz = 2 f /zx, and 2 f /yx 2 f /xy. (See a good Calculus text, e.g., Leithhold.) ii) Div Curl V = Div (Grad V ) i j = Div /x /y A B k /z C = Div ((C/y) (B/z))i + (A/z) (C/x))j + (B/x) (A/y))k) = (C/y) (B/z)) + (A/z) (C/x)) + (B/x) (A/y)) x y y 2C 2B 2A 2C 2B 2A = + + yx zx zy xy xz yz This is not zero for any vector eld V . However, Div Curl V = 0, for example, for any open ball in R3 , in which A, B, and C are all simultaneously continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 A/zy = 2 A/yz, and 2 B/xz = 2 B/zx, and 2 C/yx 2 C/xy. (See a good Calculus text, e.g., Leithhold.) 2 2. a) T = x3 + 7x2 z 2 3xy + 11xyz + 3yz + 34z + 39 T 2T = 3x2 + 14xz 2 3y + 11yz = 6x + 14z 2 x 2x T = 3x + 11xz + 3z y 2T =0 2y T = 14x2 z + 11xy + 3y + 34 z 2T = 14x2 2y b) L = 200exyz L = 200yzexyz x Grad L = L x L = 200xzexyz y i+ L y j+ L = 200xyexyz z L z = 200yzexyz i 200xzexyz j 200xyexyz k = Grad L |(1,1,0) 200(1)(0)e(1)(1)(0) i 200(1)(0)e(1)(1)(0) j 200(1)(1)e(1)(1)(0) k = 0i + 0j 200k Grad L = 200k The size of the illuminance at (1,1,0) is: Grad L |(1,1,0) = = = Grad L |(1,1,0) Grad L |(1,1,0) (200k) (200k) 2002 Grad L |(1,1,0) = 200 For this particular problem (together with the assumptions in the Note below), the size of the illuminance at a point is a summary measure of the absolute rate of decrease of the illuminance along the three coordinate axes at that particular point. (Note: We assume that the coordinate axes are tted so that the room would lie in the rst octant with (0,0,0) as one of the corners of the room. As a consequence, x 0, y 0, and z 0. This means that L decreases as we move further away from the origin inside the room (not at the edges or on the surfaces where at least one of the three coordinates is zero) since the exponent of the negative exponential would get bigger in magnitude when at least one of the coordinates increase(s) and neither one is zero. This can also be seen from rst partial derivatives of L, all of which are negative when x > 0, y > 0, and z > 0 . Hence, the illuminance L decreases with increasing coordinate be it x, or y or z.) c) T = 100xyez T = 100yez x T = 100xez y 1 T = 100xyez z Grad T = T x T y i+ j+ T z = 100yez i + 100xez j 100xyez k Grad T |(1,2,0) = 100(2)e0 i + 100(1)e0 j 100(1)(2)e0 k Grad T |(1,2,0) = 200i + 100j 200k Grad T |(1,2,0) = Grad T |(1,2,0) Grad T |(1,2,0) = (200i + 100j 200k) (200i + 100j 200k) = 2002 + 1002 + 2002 = (9)(100)2 Grad T |(1,2,0) = 300 d) = eAx+By+Cz , where A, B, and C are constants 2 = AeAx+By+Cz = A2 eAx+By+Cz x 2x = BeAx+By+Cz y 2 = B 2 eAx+By+Cz 2y 2 = CeAx+By+Cz = C 2 eAx+By+Cz z 2y Since, at (0,0,0): = 0.3 and = BeAx+By+Cz , then i) y y BeAx+By+Cz (0,0,0) = Be0+0+0 = B = 0.3 ii) = 0.4 and = CeAx+By+Cz , then z z CeAx+By+Cz (0,0,0) = Ce0+0+0 = C = 0.4 To nd A: From the Poisson's Equation, 2 2 2 + 2 + 2 = , 2x y z hence, by substituting the expressions we obtained above for the second partial derivatives of we get, A2 eAx+By+Cz + B 2 eAx+By+Cz + C 2 eAx+By+Cz = (A2 + B 2 + C 2 )eAx+By+Cz = (A2 + B 2 + C 2 ) = A2 + B 2 + C 2 = 1, 2 since = eAx+By+Cz > 0 for any (x, y, z). It follows that, A2 + (0.03)2 + (0.04)2 = 1 A2 = 1 0.09 0.16 = 0.75 A = 3/2. We are also given that, 0 as x , hence A = 3/2. If this is so, then for xed y and z, = e 3x/2+0.03y+0.04y approaches 0 as x increases without bound since its exponent will decrease without bound. Hence, = e 3x/2+0.03y+0.04y . 3. a) potential function = x2 +y 2 +z 2 ; magnetic force M = y 2 zi+z 2 xj+x2 yk i+ j+ k x y z = 2x = 2y = 2z Note that : x y z Hence, Grad = 2xi + 2yj + 2zk M M M ii) Div M = + + x y z M M M Note that : =0 =0 =0 x y z Hence, Div M = 0 + 0 + 0 = 0. iii) Div Grad i) Grad = = Div (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) (2x) (2y) (2z) + + x y z =2+2+2 = = 6. Hence Div Grad = 6. iv) Curl Grad = Curl (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) i = /x 2x = j /y 2y k /z 2z (2z) (2y) y z (2x) (2z) z x i+ = 0i + 0j + 0k = 0 Hence, Curl Grad = 0. 3 j+ (2y) (2x) z y k v) Curl M = Grad M = Grad (y 2 zi + z 2 xj + x2 yk) i = /x y2 z = j /y z2x k /z x2 y (x2 y) (z 2 x) y z (y 2 z) (x2 y) z x i+ j+ (z 2 x) (y 2 z) z y k = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k Hence, curl M = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k. vi) Div Curl M = Div (y 2 zi + z 2 xj + x2 yk) = Grad (y 2 zi + z 2 xj + x2 yk) (x2 2zx) (y 2 2xy) (z 2 2yz) + + x y z = (2x 2z) + (2y 2x) + (2z 2y) = = (2x 2x) + (2y 2y) + (2z 2z) = 0 Hence, Div Curl M = 0. b) i) Curl Grad f f f f i+ j+ k x y z f f f = Grad i+ j+ k x y z = Curl i = /x f x = j /y k /z f y f z f ( f ) ( y ) z y x ( f ) ( f ) x z z x i+ j+ ( f ) y x ( f ) x y k = ( 2 f /zy 2 f /yz)i + ( 2 f /xz 2 f /zz)j + ( 2 f /yx 2 f /xy)k This is not the zero vector for any f . However, Curl Grad f = 0, for example, in any open ball of R3 , in which f is continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 f /zy = 2 f /yz, and 2 f /xz = 2 f /zx, and 2 f /yx = 2 f /xy. (See a good Calculus text, e.g., Leithhold.) 4 ii) Div Curl V = Div (Grad V ) i j = Div /x /y A B k /z C = Div ((C/y) (B/z))i + (A/z) (C/x))j + (B/x) (A/y))k) = (C/y) (B/z)) + (A/z) (C/x)) + (B/x) (A/y)) x y y 2C 2B 2A 2C 2B 2A = + + yx zx zy xy xz yz This is not zero for any vector eld V . However, Div Curl V = 0, for example, in any open ball of R3 , in which A, B, and C are all simultaneously continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 A/zy = 2 A/yz, and 2 B/xz = 2 B/zx, and 2 C/yx = 2 C/xy. (See a good Calculus text, e.g., Leithhold.) 5 1. Given point (1,2,5) and surface z = 3x + 4y + 5, we want to nd point (x, y, z) on surface that is closest to (1,2,5), i.e., to minimize D = (x 1)2 + (y 2)2 + (z 5)2 . Since (x, y, z on surface, then D = (x 1)2 + (y 2)2 + (3x + 4y + 5 5)2 = (x 1)2 + (y 2)2 + (3x + 4y)2 = x2 2x + 1 + y 2 4y + 4 + 9x2 + 24xy + 16x2 = 10x2 + 24xy + 17y 2 2x 4y + 5 Hence, D = 20x + 24y 2 x 2D = 20 2x D = 34y + 24x 4 y 2D = 34 2y 2D = 24 xy Observe that we can already see that there is a point of relative minimun even without solving for the critical value(s) since 2D 2D 2D = (20)(34) (24)2 = 680 576 = 104 > 0 and 2x 2y xy 2D > 0. 2x To solve for the critical value(s), consider the system of equation: D = 0 = 20x + 24y 2 = 0 x D = 0 = 24x + 34y 4 = 0 y or equivalently, D = 0 = 10x + 12y = 1 x D = 0 = 12x + 17y = 2 y This is a system of two linear equations in the unknowns x and y with solutions (Cramer's Rule) 1 12 2 17 7 x= = 26 10 12 12 17 10 1 12 2 8 4 y= = = 26 13 10 12 12 17 1 4 Hence, the point of relative minimum is ( 7 , 13 ). 26 This must also be a point of absolute minimum since the given surface is a plane and, from geometry, given a point not on a plane, the shortest distance from the point to the plane exists and is the perpendicular distance from that given point to the plane. ((1,2,5) is not on z = 3x + 4y + 5 since it does not satisfy the equation of this surface). Hence, the point on the plane z = 3x+4y+5 that is closest to the point (1,2,5) 4 4 is the point with coordinates x = 7 , y = 13 , z = 3( 7 ) + 4( 13 ) + 5 = 141 or 26 26 26 7 4 141 ( 26 , 13 , 26 ). By substituion into D, one can verify that the minimum distance attained is 112626 . 2. a) T = x3 + 7x2 z 2 3xy + 11xyz + 3yz + 34z + 39 T 2T = 3x2 + 14xz 2 3y + 11yz = 6x + 14z 2 x 2x 2T =0 2y T = 3x + 11xz + 3z y 2T T = 14x2 z + 11xy + 3y + 34 = 14x2 z 2z b) L = 200exyz L L L = 200yzexyz = 200xzexyz = 200xyexyz x y z Grad L = L x i+ L y j+ L z = 200yzexyz i 200xzexyz j 200xyexyz k = Grad L |(1,1,0) 200(1)(0)e(1)(1)(0) i 200(1)(0)e(1)(1)(0) j 200(1)(1)e(1)(1)(0) k = 0i + 0j 200k Grad L = 200k The size of the gradient of the illuminance at (1,1,0) is: Grad L |(1,1,0) = = = (Grad L |(1,1,0) ) (Grad L |(1,1,0) ) (200k) (200k) 2002 Grad L |(1,1,0) = 200 For this particular problem (together with the assumptions in the Note below), the size of the gradient of the illuminance ( Grad L ) at a point, is a summary measure of the absolute rate of decrease of the illuminance along the three coordinate axes at that particular point. (Note: We assume that the coordinate axes are tted so that the room would lie in the rst octant with (0,0,0) as one of the corners of the room. As a consequence, x 0, y 0, and z 0. 2 This means that L decreases as we move further away from the origin inside the room (not at the edges or on the surfaces where at least one of the three coordinates is zero) since the exponent of the negative exponential would get bigger in magnitude when at least one of the coordinates increase(s) and neither one is zero. This can also be seen from rst partial derivatives of L, all of which are negative when x > 0, y > 0, and z > 0 . Hence, the illuminance L decreases with increasing coordinate be it x, or y or z.) c) T = 100xyez T = 100yez x T = 100xez y Grad T = T x i+ T = 100xyez z T y j+ T z = 100yez i + 100xez j 100xyez k Grad T |(1,2,0) = 100(2)e0 i + 100(1)e0 j 100(1)(2)e0 k Grad T |(1,2,0) = 200i + 100j 200k Grad T |(1,2,0) = (Grad T |(1,2,0) ) (Grad T |(1,2,0) ) = (200i + 100j 200k) (200i + 100j 200k) = 2002 + 1002 + 2002 = (9)(100)2 Grad T |(1,2,0) = 300 d) = eAx+By+Cz , where A, B, and C are constants 2 = AeAx+By+Cz = A2 eAx+By+Cz x 2x = BeAx+By+Cz y 2 = B 2 eAx+By+Cz 2y 2 = CeAx+By+Cz = C 2 eAx+By+Cz z 2z Since, at (0,0,0): i) = 0.3 and = BeAx+By+Cz , then y y BeAx+By+Cz (0,0,0) = Be0+0+0 = B = 0.3 ii) = 0.4 and = CeAx+By+Cz , then z z CeAx+By+Cz (0,0,0) = Ce0+0+0 = C = 0.4 To nd A: From the Poisson's Equation, 2 2 2 + 2 + 2 = , 2x y z 3 hence, by substituting the expressions we obtained above for the second partial derivatives of we get, A2 eAx+By+Cz + B 2 eAx+By+Cz + C 2 eAx+By+Cz = (A2 + B 2 + C 2 )eAx+By+Cz = (A2 + B 2 + C 2 ) = A2 + B 2 + C 2 = 1, since = eAx+By+Cz > 0 for any (x, y, z). It follows that, A2 + (0.03)2 + (0.04)2 = 1 A2 = 1 0.09 0.16 = 0.75 A = 3/2. We are also given that, 0 as x , hence A = 3/2. If this is so, then for xed y and z, = e 3x/2+0.03y+0.04y approaches 0 as x increases without bound since its exponent will decrease without bound. Hence, = e 3x/2+0.03y+0.04y . 3. a) potential function = x2 +y 2 +z 2 ; magnetic force M = y 2 zi+z 2 xj+x2 yk i+ j+ k x y z Note that : = 2x = 2y = 2z x y z Hence, Grad = 2xi + 2yj + 2zk M M M + + ii) Div M = x y z M M M Note that : =0 =0 =0 x y z Hence, Div M = 0 + 0 + 0 = 0. iii) Div Grad i) Grad = = Div (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) (2x) (2y) (2z) + + x y z =2+2+2 = = 6. Hence Div Grad = 6. 4 iv) Curl Grad = Curl (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) i = /x 2x = j /y 2y k /z 2z (2z) (2y) y z (2x) (2z) z x i+ (2y) (2x) z y j+ k = 0i + 0j + 0k = 0 Hence, Curl Grad = 0. v) Curl M = Grad M = Grad (y 2 zi + z 2 xj + x2 yk) i = /x y2 z = j /y z2x k /z x2 y (x2 y) (z 2 x) y z (y 2 z) (x2 y) z x i+ j+ (z 2 x) (y 2 z) z y k = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k Hence, curl M = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k. vi) Div Curl M = Div (y 2 zi + z 2 xj + x2 yk) = Grad (y 2 zi + z 2 xj + x2 yk) (x2 2zx) (y 2 2xy) (z 2 2yz) + + x y z = (2x 2z) + (2y 2x) + (2z 2y) = = (2x 2x) + (2y 2y) + (2z 2z) = 0 b) Hence, Div Curl M = 0. i) Curl Grad f f f f i+ j+ k x y z f f f = Grad i+ j+ k x y z = Curl i = /x f x = j /y k /z f y f z f ( f ) ( y ) z y x ( f ) ( f ) x z z x i+ j+ ( f ) y x ( f ) x y k = ( 2 f /zy 2 f /yz)i + ( 2 f /xz 2 f /zz)j + ( 2 f /yx 2 f /xy)k 5 This is not the zero vector for any f . However, Curl Grad f = 0, for example, in any open ball of R3 , in which f is continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 f /zy = 2 f /yz, and 2 f /xz = 2 f /zx, and 2 f /yx = 2 f /xy. (See a good Calculus text, e.g., Leithhold.) ii) Div Curl V = Div (Grad V ) i j = Div /x /y A B k /z C = Div ((C/y) (B/z))i + (A/z) (C/x))j + (B/x) (A/y))k) = (C/y) (B/z)) + (A/z) (C/x)) + (B/x) (A/y)) x y y 2B 2A 2C 2B 2A 2C + + = yx zx zy xy xz yz This is not zero for any vector eld V . However, Div Curl V = 0, for example, in any open ball of R3 , in which A, B, and C are all simultaneously continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 A/zy = 2 A/yz, and 2 B/xz = 2 B/zx, and 2 C/yx = 2 C/xy. (See a good Calculus text, e.g., Leithhold.) 6 1. Given point (1,2,5) and surface z = 3x+4y+5, we want to nd the point (x, y, z) on the given surface that is closest to (1,2,5), i.e., to minimize D = (x1)2 +(y2)2 +(z5)2 . Since (x, y, z) is on the surface, then D can be written as a function of two variables, x, and y: D = (x 1)2 + (y 2)2 + (3x + 4y + 5 5)2 = (x 1)2 + (y 2)2 + (3x + 4y)2 = x2 2x + 1 + y 2 4y + 4 + 9x2 + 24xy + 16x2 = 10x2 + 24xy + 17y 2 2x 4y + 5 Consequently, D = 20x + 24y 2 x 2D = 20 2x D = 34y + 24x 4 y 2D = 34 2y 2D = 24 xy () Observe that from the above information, we can already see that any critical point that is a point of relative extremum must be a point of relative minimum even without solving for the critical point(s) since the second partial derivatives in this problem are all constants and by the Second Partial Derivative Test: 2D 2D 2D = (20)(34) (24)2 = 680 576 = 104 > 0 and 2x 2y xy 2D > 0. 2x To solve for the critical point(s), consider the system of equations obtained by equating the rst partial derivatives to 0: D = 0 = 20x + 24y 2 = 0 x D = 0 = 24x + 34y 4 = 0 y or equivalently, D = 0 = 10x + 12y = 1 x D = 0 = 12x + 17y = 2. y This is a system of two linear equations in the unknowns x and y with solutions 1 (Cramer's Rule) 1 12 2 17 7 x= = 26 10 12 12 17 10 1 12 2 8 4 = y= = 26 13 10 12 12 17 4 Hence, the only possible point of relative extremum is ( 7 , 13 ) and it is a point of 26 relative minimum (see () above). Now, D is a polynomial in x and y and therefore must be continuous on all of R2 . 4 Hence, ( 7 , 13 ) is the point of absolute minimum of D in R2 , in the sense that this 26 point is the only point of relative minimum of D in any open disk in R2 , no matter how large, that contains this point. Thus, it must be the point of absolute minimum of D on any such disk, no matter how large. (To re-iterate the fact that the point 4 gives a relative minimum: On any open disk in R2 that contains ( 7 , 13 ), this point 26 is the only point of relative extremum of D, since this is the only point at which the rst partial derivatives of D evaluate to 0 (critical point), and gives a point of relative minimum of D, as can be seen in the Second Partial Derivatives Test). There must also be a point of absolute minimum since the given surface is a plane and, from geometry, given a point not on a plane, the shortest distance from the point to the plane exists and is the perpendicular distance from that given point to the plane. ((1,2,5) is not on z = 3x + 4y + 5 since it does not satisfy the equation of this surface). Hence, the point on the plane z = 3x + 4y + 5 that is closest to the point (1,2,5) is the 4 4 4 point with coordinates x = 7 , y = 13 , z = 3( 7 ) + 4( 13 ) + 5 = 141 or ( 7 , 13 , 141 ). 26 26 26 26 26 By substituion into D, one can verify that the minimum distance attained is 112626 . (The same answer is obtained when vector analysis is used verifying that the point obtained is indeed the point of absolute minimum.) 2. a) T = x3 + 7x2 z 2 3xy + 11xyz + 3yz + 34z + 39 T 2T = 3x2 + 14xz 2 3y + 11yz = 6x + 14z 2 x 2x T = 3x + 11xz + 3z y 2T =0 2y T = 14x2 z + 11xy + 3y + 34 z 2T = 14x2 2z 2 b) L = 200exyz L = 200yzexyz x Grad L = L x L = 200xzexyz y i+ L y j+ L = 200xyexyz z L z = 200yzexyz i 200xzexyz j 200xyexyz k = Grad L |(1,1,0) 200(1)(0)e(1)(1)(0) i 200(1)(0)e(1)(1)(0) j 200(1)(1)e(1)(1)(0) k = 0i + 0j 200k Grad L = 200k The size of the gradient of the illuminance at (1,1,0) is: Grad L |(1,1,0) = = = (Grad L |(1,1,0) ) (Grad L |(1,1,0) ) (200k) (200k) 2002 Grad L |(1,1,0) = 200 For this particular problem (together with the assumptions in the Note below), the size of the gradient of the illuminance ( Grad L ) at a point, is a summary measure of the absolute rate of decrease of the illuminance along the three coordinate axes at that particular point. (Note: We assume that the coordinate axes are tted so that the room would lie in the rst octant with (0,0,0) as one of the corners of the room. As a consequence, x 0, y 0, and z 0. This means that L decreases as we move further away from the origin inside the room (not at the edges or on the surfaces where at least one of the three coordinates is zero) since the exponent of the negative exponential would get bigger in absolute magnitude when at least one of the coordinates increases and neither one is zero. This can also be seen from the rst partial derivatives of L, all of which are negative when x > 0, y > 0, and z > 0 . Hence, the illuminance L decreases with increasing coordinate be it x, or y or z.) c) T = 100xyez T = 100yez x T = 100xez y T = 100xyez z 3 Grad T = T x i+ T y j+ T z = 100yez i + 100xez j 100xyez k Grad T |(1,2,0) = 100(2)e0 i + 100(1)e0 j 100(1)(2)e0 k Grad T |(1,2,0) = 200i + 100j 200k Grad T |(1,2,0) = (Grad T |(1,2,0) ) (Grad T |(1,2,0) ) = (200i + 100j 200k) (200i + 100j 200k) = 2002 + 1002 + 2002 = (9)(100)2 Grad T |(1,2,0) = 300 d) = eAx+By+Cz , where A, B, and C are constants 2 = AeAx+By+Cz = A2 eAx+By+Cz x 2x = BeAx+By+Cz y 2 = B 2 eAx+By+Cz 2y 2 = CeAx+By+Cz = C 2 eAx+By+Cz z 2z Since, at (0,0,0): = 0.3 and = BeAx+By+Cz , then i) y y BeAx+By+Cz (0,0,0) = Be0+0+0 = B = 0.3 ii) = 0.4 and = CeAx+By+Cz , then z z CeAx+By+Cz (0,0,0) = Ce0+0+0 = C = 0.4 To nd A: From the Poisson's Equation, 2 2 2 + 2 + 2 = , 2x y z hence, by substituting the expressions we obtained above for the second partial derivatives of we get, A2 eAx+By+Cz + B 2 eAx+By+Cz + C 2 eAx+By+Cz = (A2 + B 2 + C 2 )eAx+By+Cz = (A2 + B 2 + C 2 ) = A2 + B 2 + C 2 = 1, 4 since = eAx+By+Cz > 0 for any (x, y, z). It follows that, A2 + (0.03)2 + (0.04)2 = 1 A2 = 1 0.09 0.16 = 0.75 A = 3/2. We are also given that, 0 as x , hence A = 3/2. If this is so, then for xed y and z, = e 3x/2+0.03y+0.04y approaches 0 as x increases without bound since its exponent will decrease without bound. Hence, = e 3x/2+0.03y+0.04y . 3. a) potential function = x2 + y 2 + z 2 ; magnetic force M = y 2 zi + z 2 xj + x2 yk i+ j+ k x y z = 2x = 2y = 2z Note that : x y z Hence, Grad = 2xi + 2yj + 2zk. M M M + + ii) Div M = x y z M M M Note that : =0 =0 =0 x y z Hence, Div M = 0 + 0 + 0 = 0. iii) Div Grad i) Grad = = Div (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) (2x) (2y) (2z) + + x y z =2+2+2 = = 6. Hence Div Grad = 6. iv) Curl Grad = Curl (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) i = /x 2x = j /y 2y k /z 2z (2z) (2y) y z i+ (2x) (2z) z x = 0i + 0j + 0k = 0 Hence, Curl Grad = 0. 5 j+ (2y) (2x) z y k v) Curl M = Grad M = Grad (y 2 zi + z 2 xj + x2 yk) i = /x y2 z = j /y z2x k /z x2 y (x2 y) (z 2 x) y z (y 2 z) (x2 y) z x i+ j+ (z 2 x) (y 2 z) z y k = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k Hence, curl M = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k. vi) Div Curl M = Div (y 2 zi + z 2 xj + x2 yk) = Grad (y 2 zi + z 2 xj + x2 yk) (x2 2zx) (y 2 2xy) (z 2 2yz) + + x y z = (2x 2z) + (2y 2x) + (2z 2y) = = (2x 2x) + (2y 2y) + (2z 2z) = 0 Hence, Div Curl M = 0. b) i) Curl Grad f f f f i+ j+ k x y z f f f = Grad i+ j+ k x y z = Curl i = /x f x = j /y k /z f y f z f ( f ) ( y ) z y x i+ ( f ) ( f ) x z z x j+ ( f ) y x ( f ) x y k = ( 2 f /zy 2 f /yz)i + ( 2 f /xz 2 f /zz)j + ( 2 f /yx 2 f /xy)k This is not the zero vector for any f . However, Curl Grad f = 0, for example, in any open ball of R3 , in which f is continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 f /zy = 2 f /yz, and 2 f /xz = 2 f /zx, and 2 f /yx = 2 f /xy. (See a good Calculus text, e.g., Leithhold.) 6 ii) Div Curl V = Div (Grad V ) i j = Div /x /y A B k /z C = Div ((C/y) (B/z))i + (A/z) (C/x))j + (B/x) (A/y))k) = (C/y) (B/z)) + (A/z) (C/x)) + (B/x) (A/y)) x y y 2C 2B 2A 2C 2B 2A = + + yx zx zy xy xz yz This is not zero for any vector eld V . However, Div Curl V = 0, for example, in any open ball of R3 , in which A, B, and C are all simultaneously continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 A/zy = 2 A/yz, and 2 B/xz = 2 B/zx, and 2 C/yx = 2 C/xy. (See a good Calculus text, e.g., Leithhold.) 7 1. Given the point (1,2,5) and the surface z = 3x + 4y + 5, we want to nd the point (x, y, z) on the given surface that is closest to (1,2,5), i.e., to minimize D = (x 1)2 + (y 2)2 + (z 5)2 . Since (x, y, z) is on the surface, then D can be written as a function of two variables, x and y: D = (x 1)2 + (y 2)2 + (z 5)2 = (x 1)2 + (y 2)2 + ((3x + 4y + 5) 5)2 = (x 1)2 + (y 2)2 + (3x + 4y)2 = x2 2x + 1 + y 2 4y + 4 + 9x2 + 24xy + 16x2 = 10x2 + 24xy + 17y 2 2x 4y + 5 Consequently, D = 20x + 24y 2 x 2D = 20 2x D = 34y + 24x 4 y 2D = 34 2y 2D = 24 xy () Observe that from the above information, we can already see that any critical point that is a point of relative extremum must be a point of relative minimum even without solving for the critical point(s) since the second partial derivatives in this problem are all constants and by the Second-Derivative Test: 2D 2D 2D = (20)(34) (24)2 = 680 576 = 104 > 0 and 2x 2y xy 2D > 0. 2x To solve for the critical point(s), consider the system of equations obtained by equating the rst partial derivatives to 0: D = 0 = 20x + 24y 2 = 0 x D = 0 = 24x + 34y 4 = 0 y or equivalently, D = 0 = 10x + 12y = 1 x D = 0 = 12x + 17y = 2. y This is a system of two linear equations in the unknowns x and y with solutions 1 (Cramer's Rule) 1 12 2 17 7 x= = 26 10 12 12 17 10 1 12 2 8 4 = y= = 26 13 10 12 12 17 4 Hence, the only possible point of relative extremum is ( 7 , 13 ) and it is a point of 26 relative minimum (see () above). Now, D is a polynomial in x and y and therefore must be continuous on all of R2 . 4 Hence, ( 7 , 13 ) is the point of absolute minimum of D in R2 , in the sense that this 26 point is the only point of relative minimum of D in any open disk in R2 , no matter how large, that contains this point. Thus, it must be the point of absolute minimum of D on any such disk, no matter how large. (To re-iterate the fact that the point 4 gives a relative minimum: On any open disk in R2 that contains ( 7 , 13 ), this point 26 is the only point of relative extremum of D, since this is the only point at which the rst partial derivatives of D evaluate to 0 (critical point), and gives a point of relative minimum of D, as can be seen in the Second-Derivative Test). There must also be a point of absolute minimum since the given surface is a plane and, from geometry, given a point not on a plane, the shortest distance from the point to the plane exists and is the perpendicular distance from that given point to the plane. ((1,2,5) is not on z = 3x + 4y + 5 since it does not satisfy the equation of this surface). Hence, the point on the plane z = 3x + 4y + 5 that is closest to the point (1,2,5) is the 4 4 4 point with coordinates x = 7 , y = 13 , z = 3( 7 ) + 4( 13 ) + 5 = 141 or ( 7 , 13 , 141 ). 26 26 26 26 26 By substituion into D, one can verify that the minimum distance attained is 112626 . (The same answer is obtained when vector analysis is used verifying that the point obtained is indeed the point of absolute minimum.) 2. a) T = x3 + 7x2 z 2 3xy + 11xyz + 3yz + 34z + 39 T 2T = 3x2 + 14xz 2 3y + 11yz = 6x + 14z 2 x 2x T = 3x + 11xz + 3z y 2T =0 2y T = 14x2 z + 11xy + 3y + 34 z 2T = 14x2 2z 2 b) L = 200exyz L = 200yzexyz x Grad L = L x L = 200xzexyz y i+ L y j+ L = 200xyexyz z L z = 200yzexyz i 200xzexyz j 200xyexyz k = Grad L |(1,1,0) 200(1)(0)e(1)(1)(0) i 200(1)(0)e(1)(1)(0) j 200(1)(1)e(1)(1)(0) k = 0i + 0j 200k Grad L = 200k The size of the gradient of the illuminance at (1,1,0) is: Grad L |(1,1,0) = = = (Grad L |(1,1,0) ) (Grad L |(1,1,0) ) (200k) (200k) 2002 Grad L |(1,1,0) = 200 For this particular problem (together with the assumptions in the Note below), the size of the gradient of the illuminance ( Grad L ) at a point, is a summary measure of the absolute rate of decrease of the illuminance along the three coordinate axes at that particular point. (Note: We assume that the coordinate axes are tted so that the room would lie in the rst octant with (0,0,0) as one of the corners of the room. As a consequence, x 0, y 0, and z 0. This means that L decreases as we move further away from the origin inside the room (not at the edges or on the surfaces where at least one of the three coordinates is zero) since the exponent of the negative exponential would get bigger in absolute magnitude when at least one of the coordinates increases and neither one is zero. This can also be seen from the rst partial derivatives of L, all of which are negative when x > 0, y > 0, and z > 0 . Hence, the illuminance L decreases with increasing coordinate be it x, or y or z.) c) T = 100xyez T = 100yez x T = 100xez y T = 100xyez z 3 Grad T = T x i+ T y j+ T z = 100yez i + 100xez j 100xyez k Grad T |(1,2,0) = 100(2)e0 i + 100(1)e0 j 100(1)(2)e0 k Grad T |(1,2,0) = 200i + 100j 200k Grad T |(1,2,0) = (Grad T |(1,2,0) ) (Grad T |(1,2,0) ) = (200i + 100j 200k) (200i + 100j 200k) = 2002 + 1002 + 2002 = (9)(100)2 Grad T |(1,2,0) = 300 d) = eAx+By+Cz , where A, B, and C are constants 2 = AeAx+By+Cz = A2 eAx+By+Cz x 2x = BeAx+By+Cz y 2 = B 2 eAx+By+Cz 2y 2 = CeAx+By+Cz = C 2 eAx+By+Cz z 2z Since, at (0,0,0): = 0.3 and = BeAx+By+Cz , then i) y y BeAx+By+Cz (0,0,0) = Be0+0+0 = B = 0.3 ii) = 0.4 and = CeAx+By+Cz , then z z CeAx+By+Cz (0,0,0) = Ce0+0+0 = C = 0.4 To nd A: From the Poisson's Equation, 2 2 2 + 2 + 2 = , 2x y z hence, by substituting the expressions we obtained above for the second partial derivatives of we get, A2 eAx+By+Cz + B 2 eAx+By+Cz + C 2 eAx+By+Cz = (A2 + B 2 + C 2 )eAx+By+Cz = (A2 + B 2 + C 2 ) = A2 + B 2 + C 2 = 1, 4 since = eAx+By+Cz > 0 for any (x, y, z). It follows that, A2 + (0.03)2 + (0.04)2 = 1 A2 = 1 0.09 0.16 = 0.75 A = 3/2. We are also given that, 0 as x , hence A = 3/2. If this is so, then for xed y and z, = e 3x/2+0.03y+0.04y approaches 0 as x increases without bound since its exponent will decrease without bound. Hence, = e 3x/2+0.03y+0.04y . 3. a) potential function = x2 + y 2 + z 2 ; i) Grad = i+ j+ k x y z Note that : = 2x = 2y x y Hence, Grad = 2xi + 2yj + 2zk. M M M ii) Div M = + + x y z M M Note that : =0 =0 x y Hence, Div M = 0 + 0 + 0 = 0. iii) Div Grad magnetic force M = y 2 zi + z 2 xj + x2 yk = 2z z M =0 z = Div (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) (2x) (2y) (2z) + + x y z =2+2+2 = = 6. Hence Div Grad = 6. iv) Curl Grad = Curl (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) i = /x 2x = j /y 2y k /z 2z (2z) (2y) y z i+ (2x) (2z) z x = 0i + 0j + 0k = 0 Hence, Curl Grad = 0. 5 j+ (2y) (2x) z y k v) Curl M = Grad M = Grad (y 2 zi + z 2 xj + x2 yk) i = /x y2 z = j /y z2x k /z x2 y (x2 y) (z 2 x) y z (y 2 z) (x2 y) z x i+ j+ (z 2 x) (y 2 z) z y k = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k Hence, Curl M = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k. vi) Div Curl M = Div (y 2 zi + z 2 xj + x2 yk) = Grad (y 2 zi + z 2 xj + x2 yk) (x2 2zx) (y 2 2xy) (z 2 2yz) + + x y z = (2x 2z) + (2y 2x) + (2z 2y) = = (2x 2x) + (2y 2y) + (2z 2z) = 0 Hence, Div Curl M = 0. b) i) Curl Grad f f f f i+ j+ k x y z f f f = Grad i+ j+ k x y z = Curl i = /x f x = j /y k /z f y f z f ( f ) ( y ) z y x i+ ( f ) ( f ) x z z x j+ ( f ) y x ( f ) x y k = ( 2 f /zy 2 f /yz)i + ( 2 f /xz 2 f /zz)j + ( 2 f /yx 2 f /xy)k This is not the zero vector for any f . However, Curl Grad f = 0, for example, in any open ball of R3 , in which f is continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 f /zy = 2 f /yz, and 2 f /xz = 2 f /zx, and 2 f /yx = 2 f /xy. (See a good Calculus text, e.g., Leithhold.) 6 ii) Div Curl V = Div (Grad V ) i j = Div /x /y A B k /z C = Div ((C/y) (B/z))i + (A/z) (C/x))j + (B/x) (A/y))k) = (C/y) (B/z)) + (A/z) (C/x)) + (B/x) (A/y)) x y y 2C 2B 2A 2C 2B 2A = + + yx zx zy xy xz yz This is not zero for any vector eld V . However, Div Curl V = 0, for example, in any open ball of R3 , in which A, B, and C are all simultaneously continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 A/zy = 2 A/yz, and 2 B/xz = 2 B/zx, and 2 C/yx = 2 C/xy. (See a good Calculus text, e.g., Leithhold.) 7 1. Given the point (1,2,5) and the surface z = 3x + 4y + 5, we want to nd the point (x, y, z) on the given surface that is closest to (1,2,5), i.e., to minimize D = (x 1)2 + (y 2)2 + (z 5)2 . Since (x, y, z) is on the surface, then D can be written as a function of two variables, x and y: D = (x 1)2 + (y 2)2 + (z 5)2 = (x 1)2 + (y 2)2 + ((3x + 4y + 5) 5)2 = (x 1)2 + (y 2)2 + (3x + 4y)2 = x2 2x + 1 + y 2 4y + 4 + 9x2 + 24xy + 16x2 = 10x2 + 24xy + 17y 2 2x 4y + 5 Consequently, D = 20x + 24y 2 x 2D = 20 2x D = 34y + 24x 4 y 2D = 34 2y 2D = 24 xy () Observe that from the above information, we can already see that any critical point that is a point of relative extremum must be a point of relative minimum even without solving for the critical point(s) since the second partial derivatives in this problem are all constants and by the Second-Derivative Test: 2D 2D 2D = (20)(34) (24)2 = 680 576 = 104 > 0 and 2x 2y xy 2D > 0. 2x To solve for the critical point(s), consider the system of equations obtained by equating the rst partial derivatives to 0: D = 0 = 20x + 24y 2 = 0 x D = 0 = 24x + 34y 4 = 0 y or equivalently, D = 0 = 10x + 12y = 1 x D = 0 = 12x + 17y = 2. y This is a system of two linear equations in the unknowns x and y with solutions 1 (Cramer's Rule) 1 12 2 17 7 x= = 26 10 12 12 17 10 1 12 2 8 4 = y= = 26 13 10 12 12 17 4 Hence, the only possible point of relative extremum is ( 7 , 13 ) and it is a point of 26 relative minimum (see () above). Now, D is a polynomial in x and y and therefore must be continuous on all of R2 . 4 Hence, ( 7 , 13 ) is the point of absolute minimum of D in R2 , in the sense that this 26 point is the only point of relative minimum of D in any open disk in R2 , no matter how large, that contains this point. Thus, it must be the point of absolute minimum of D on any such disk, no matter how large. (To re-iterate the fact that the point 4 gives a relative minimum: On any open disk in R2 that contains ( 7 , 13 ), this point 26 is the only point of relative extremum of D, since this is the only point at which the rst partial derivatives of D evaluate to 0 (critical point), and gives a point of relative minimum of D, as can be seen in the Second-Derivative Test). There must also be a point of absolute minimum since the given surface is a plane and, from geometry, given a point not on a plane, the shortest distance from the point to the plane exists and is the perpendicular distance from that given point to the plane. ((1,2,5) is not on z = 3x + 4y + 5 since it does not satisfy the equation of this surface). Hence, the point on the plane z = 3x + 4y + 5 that is closest to the point (1,2,5) is the 4 4 4 point with coordinates x = 7 , y = 13 , z = 3( 7 ) + 4( 13 ) + 5 = 141 or ( 7 , 13 , 141 ). 26 26 26 26 26 By substituion into D, one can verify that the minimum distance attained is 112626 . (The same answer is obtained when vector analysis is used verifying that the point obtained is indeed the point of absolute minimum.) 2. a) T = x3 + 7x2 z 2 3xy + 11xyz + 3yz + 34z + 39 T 2T = 3x2 + 14xz 2 3y + 11yz = 6x + 14z 2 x 2x T = 3x + 11xz + 3z y 2T =0 2y T = 14x2 z + 11xy + 3y + 34 z 2T = 14x2 2z 2 b) L = 200exyz L = 200yzexyz x Grad L = L x L = 200xzexyz y i+ L y j+ L = 200xyexyz z L z = 200yzexyz i 200xzexyz j 200xyexyz k = Grad L |(1,1,0) 200(1)(0)e(1)(1)(0) i 200(1)(0)e(1)(1)(0) j 200(1)(1)e(1)(1)(0) k = 0i + 0j 200k Grad L = 200k The size of the gradient of the illuminance at (1,1,0) is: Grad L |(1,1,0) = = = (Grad L |(1,1,0) ) (Grad L |(1,1,0) ) (200k) (200k) 2002 Grad L |(1,1,0) = 200 For this particular problem (together with the assumptions in the Note below), the size of the gradient of the illuminance ( Grad L ) at a point, is a summary measure of the absolute rate of decrease of the illuminance along the three coordinate axes at that particular point. (Note: We assume that the coordinate axes are tted so that the room would lie in the rst octant with (0,0,0) as one of the corners of the room. As a consequence, x 0, y 0, and z 0. This means that L decreases as we move further away from the origin inside the room (not at the edges or on the surfaces where at least one of the three coordinates is zero) since the exponent of the negative exponential would get bigger in absolute magnitude when at least one of the coordinates increases and neither one is zero. This can also be seen from the rst partial derivatives of L, all of which are negative when x > 0, y > 0, and z > 0 . Hence, the illuminance L decreases with increasing coordinate be it x, or y or z.) c) T = 100xyez T = 100yez x T = 100xez y T = 100xyez z 3 Grad T = T x i+ T y j+ T z = 100yez i + 100xez j 100xyez k Grad T |(1,2,0) = 100(2)e0 i + 100(1)e0 j 100(1)(2)e0 k Grad T |(1,2,0) = 200i + 100j 200k Grad T |(1,2,0) = (Grad T |(1,2,0) ) (Grad T |(1,2,0) ) = (200i + 100j 200k) (200i + 100j 200k) = 2002 + 1002 + 2002 = (9)(100)2 Grad T |(1,2,0) = 300 d) = eAx+By+Cz , where A, B, and C are constants 2 = AeAx+By+Cz = A2 eAx+By+Cz x 2x = BeAx+By+Cz y 2 = B 2 eAx+By+Cz 2y 2 = CeAx+By+Cz = C 2 eAx+By+Cz z 2z Since, at (0,0,0): = 0.3 and = BeAx+By+Cz , then i) y y BeAx+By+Cz (0,0,0) = Be0+0+0 = B = 0.3 ii) = 0.4 and = CeAx+By+Cz , then z z CeAx+By+Cz (0,0,0) = Ce0+0+0 = C = 0.4 To nd A: From the Poisson's Equation, 2 2 2 + 2 + 2 = , 2x y z hence, by substituting the expressions we obtained above for the second partial derivatives of we get, A2 eAx+By+Cz + B 2 eAx+By+Cz + C 2 eAx+By+Cz = (A2 + B 2 + C 2 )eAx+By+Cz = (A2 + B 2 + C 2 ) = A2 + B 2 + C 2 = 1, 4 since = eAx+By+Cz > 0 for any (x, y, z). It follows that, A2 + (0.03)2 + (0.04)2 = 1 A2 = 1 0.09 0.16 = 0.75 A = 3/2. We are also given that, 0 as x , hence A = 3/2. If this is so, then for xed y and z, = e 3x/2+0.03y+0.04y approaches 0 as x increases without bound since its exponent will decrease without bound. Hence, = e 3x/2+0.03y+0.04y . 3. a) potential function = x2 + y 2 + z 2 ; i) Grad = i+ j+ k x y z Note that : = 2x = 2y x y Hence, Grad = 2xi + 2yj + 2zk. M M M ii) Div M = + + x y z M M Note that : =0 =0 x y Hence, Div M = 0 + 0 + 0 = 0. iii) Div Grad magnetic force M = y 2 zi + z 2 xj + x2 yk = 2z z M =0 z = Div (2xi + 2yj + 2zk) see i) = Grad (2xi + 2yj + 2zk) (2x) (2y) (2z) + + x y z =2+2+2 = = 6. Hence Div Grad = 6. iv) Curl Grad = Curl (2xi + 2yj + 2zk) see i) (1) = Grad (2xi + 2yj + 2zk) i = /x 2x = j /y 2y k /z 2z (2z) (2y) y z i+ (2x) (2z) z x = 0i + 0j + 0k = 0 Hence, Curl Grad = 0. 5 j+ (2y) (2x) z y k v) Curl M = Grad M = Grad (y 2 zi + z 2 xj + x2 yk) i = /x y2 z = j /y z2x k /z x2 y (x2 y) (z 2 x) y z (y 2 z) (x2 y) z x i+ j+ (z 2 x) (y 2 z) z y k = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k Hence, Curl M = (x2 2zx)i + (y 2 2xy)j + (z 2 2yz)k. vi) Div Curl M = Div (y 2 zi + z 2 xj + x2 yk) 2 2 (2) 2 = Grad (y zi + z xj + x yk) (x2 2zx) (y 2 2xy) (z 2 2yz) + + x y z = (2x 2z) + (2y 2x) + (2z 2y) = = (2x 2x) + (2y 2y) + (2z 2z) = 0 Hence, Div Curl M = 0. b) i) Curl Grad f f f f i+ j+ k x y z f f f = Grad i+ j+ k x y z = Curl i = /x f x = j /y k /z f y f z f ( f ) ( y ) z y x i+ ( f ) ( f ) x z z x j+ ( f ) y x ( f ) x y k = ( 2 f /zy 2 f /yz)i + ( 2 f /xz 2 f /zz)j + ( 2 f /yx 2 f /xy)k This is not the zero vector for any f . However, Curl Grad f = 0, for example, in any open ball of R3 , in which f is continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 f /zy = 2 f /yz, and 2 f /xz = 2 f /zx, and 2 f /yx = 2 f /xy and the conclusion follows (see a good Calculus text, e.g., Leithhold fo reference). The function in Question 3.a above is such an f as can be seen from the answer to Question 3.a.iv). 6 ii) Div Curl V = Div (Grad V ) i j = Div /x /y A B k /z C = Div ((C/y) (B/z))i + (A/z) (C/x))j + (B/x) (A/y))k) = (C/y) (B/z)) + (A/z) (C/x)) + (B/x) (A/y)) x y y 2C 2B 2A 2C 2B 2A = + + yx zx zy xy xz yz 2C 2C 2A 2A 2B 2B = + + yx xy zy yz xz zx This is not zero for any vector eld V = Ai + Bj + Ck. However, Div Curl V = 0, for example, in any open ball of R3 , in which A, B, and C are all simultaneously continuous. This condition is sucient for disregarding the order in which the partial derivatives are taken so that 2 A/zy = 2 A/yz, and 2 B/xz = 2 B/zx, and 2 C/yx = 2 C/xy and the conclusion follows (see a good Calculus text, e.g., Leithhold). The magnetic force M in Question 3.a above is such a vector eld V as can be seen from the answer to Question 3.a.vi). 7