Tutorial - Vectors.pdf - Problem 3.12 Two lines in the x–y plane are described by the expressions x 2y = −6 3x 4y = 8 Line 1 Line 2 Use vector

Tutorial - Vectors.pdf - Problem 3.12 Two lines in the...

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Unformatted text preview: Problem 3.12 Two lines in the x–y plane are described by the expressions: x + 2y = −6, 3x + 4y = 8. Line 1 Line 2 Use vector algebra to find the smaller angle between the lines at their intersection point. 30 25 20 15 10 (0, 2) -35 -30 -25 -20 -15 -10 10 15 20 25 30 35 (0, -3) -10 B A (20, -13) -15 -20 -25 θAB -30 Figure P3.12: Lines 1 and 2. Solution: Intersection point is found by solving the two equations simultaneously: −2x − 4y = 12, 3x + 4y = 8. The sum gives x = 20, which, when used in the first equation, gives y = −13. Hence, intersection point is (20, −13). Another point on line 1 is x = 0, y = −3. Vector A from (0, −3) to (20, −13) is A = xˆ (20) + yˆ (−13 + 3) = xˆ 20 − yˆ 10, p √ |A| = 202 + 102 = 500. A point on line 2 is x = 0, y = 2. Vector B from (0, 2) to (20, −13) is B = xˆ (20) + yˆ (−13 − 2) = xˆ 20 − yˆ 15, |B| = p √ 202 + 152 = 625. Angle between A and B is µ µ ¶ ¶ 400 + 150 A·B −1 −1 √ √ = cos θAB = cos = 10.3◦ . |A||B| 500 · 625 Problem 3.46 For the vector field E = xˆ xz − yˆ yz2 − zˆ xy, verify the divergence theorem by computing: (a) the total outward flux flowing through the surface of a cube centered at the origin and with sides equal to 2 units each and parallel to the Cartesian axes, and (b) the integral of ∇ · E over the cube’s volume. Solution: (a) For a cube, the closed surface integral has 6 sides: Z E · ds = Ftop + Fbottom + Fright + Fleft + Ffront + Fback , n Ftop = Z 1 Z 1 x=−1 y=−1 =− Z 1 Z 1 ¡ ¢¯ xˆ xz − yˆ yz2 − zˆ xy ¯z=1 · (ˆz dy dx) xy dy dx = x=−1 y=−1 Fbottom = Z 1 Z 1 = Z 1 Z 1 Z 1 Z 1 x=−1 y=−1 õ !¯1 ¶¯1 ¯ x2 y2 ¯¯ ¯ ¯ ¯ 4 y=−1 ¯ = 0, x=−1 ¡ ¢¯ xˆ xz − yˆ yz2 − zˆ xy ¯z=−1 · (−ˆz dy dx) xy dy dx = x=−1 y=−1 õ !¯1 ¶¯1 ¯ x2 y2 ¯¯ ¯ ¯ 4 ¯y=−1 ¯ = 0, x=−1 ¢¯ Fright = xˆ xz − yˆ yz2 − zˆ xy ¯y=1 · (ˆy dz dx) x=−1 z=−1 à µ ¶¯1 !¯1 Z 1 Z 1 ¯ xz3 ¯¯ ¯ 2 =− z dz dx = − ¯ ¯ 3 x=−1 z=−1 z=−1 ¯ ¡ x=−1 Z 1 Z 1 Ffront = Z 1 Z 1 = Z 1 Z 1 = −4 , 3 ¢¯ Fleft = xˆ xz − yˆ yz2 − zˆ xy ¯y=−1 · (−ˆy dz dx) x=−1 z=−1 à µ ¶¯1 !¯1 Z 1 Z 1 ¯ −4 xz3 ¯¯ ¯ = z2 dz dx = − =− , ¯ ¯ 3 3 x=−1 z=−1 z=−1 ¯ ¡ x=−1 y=−1 z=−1 y=−1 z=−1 ¡ ¢¯ xˆ xz − yˆ yz2 − zˆ xy ¯x=1 · (ˆx dz dy) z dz dy = õ ¶¯1 !¯¯1 yz2 ¯¯ ¯ ¯ 2 ¯z=−1 ¯ y=−1 = 0, Fback = Z 1 Z 1 = Z 1 Z 1 y=−1 z=−1 ¡ ¢¯ xˆ xz − yˆ yz2 − zˆ xy ¯x=−1 · (−ˆx dz dy) z dz dy = y=−1 z=−1 Z E · ds = 0 + 0 + n õ ¶¯1 !¯¯1 yz2 ¯¯ ¯ ¯ ¯ 2 z=−1 ¯ = 0, y=−1 −8 −4 −4 + +0+0 = . 3 3 3 (b) ZZZ ∇·E dv = = Z 1 Z 1 Z 1 x=−1 y=−1 z=−1 Z 1 Z 1 Z 1 x=−1 y=−1 z=−1 à = ∇·(ˆxxz − yˆ yz2 − zˆ xy) dz dy dx (z − z2 ) dz dy dx ¯1 ¯ ¯ ¯ ¯ y=−1 ¯ ¶¶¯1 !¯¯1 µ µ 2 ¯ z3 z ¯ ¯ − xy ¯ ¯ 2 3 z=−1 ¯ x=−1 = −8 . 3 2 2 Problem Z 3.50 For the vector field E = xˆ xy − yˆ (x + 2y ), calculate (a) E · dl around the triangular contour shown in Fig. P3.50(a), and n ZC (b) S × E) · ds over the area of the triangle. (∇× Solution: In addition to the independent condition that z = 0, the three lines of the triangle are represented by the equations y = 0, x = 1, and y = x, respectively. y 1 y 1 L3 L2 0 L1 1 (a) x 0 L3 L2 L1 1 (b) 2 x Figure P3.50: Contours for (a) Problem 3.50 and (b) Problem 3.51. (a) Z E · dl = L1 + L2 + L3 , n L1 = = Z (ˆxxy − yˆ (x2 + 2y2 )) · (ˆx dx + yˆ dy + zˆ dz) Z 1 x=0 L2 = = Z (xy)|y=0,z=0 dx − Z 0 ¡ Z 0 ¢¯ (0)|y=0 dz = 0, x2 + 2y2 ¯z=0 dy + y=0 z=0 (ˆxxy − yˆ (x2 + 2y2 )) · (ˆx dx + yˆ dy + zˆ dz) Z 1 x=1 (xy)|z=0 dx − Z 1 ¡ Z 0 ¢¯ (0)|x=1 dz x2 + 2y2 ¯x=1,z=0 dy + y=0 z=0 ¶¯1 µ −5 2y3 ¯¯ , +0 = = 0− y+ ¯ 3 3 y=0 L3 = Z (ˆxxy − yˆ (x2 + 2y2 )) · (ˆx dx + yˆ dy + zˆ dz) µ ¶¯0 ¡ ¢¯0 x3 ¯¯ 2 − y3 ¯y=1 + 0 = . ¯ 3 x=1 3 Z 0 ¡ Z 0 ¢¯ 2 2 ¯ (xy)|y=x, z=0 dx − (0)|y=x dz x + 2y x=y, z=0 dy + = y=1 x=1 z=0 Z 0 = Therefore, 5 2 E · dl = 0 − + = −1. 3 3 (b) From Eq. (3.105), ∇×E = −ˆz3x, so that Z n ZZ ∇×E · ds = Z 1 Z x ((−ˆz3x) · (ˆz dy dx))|z=0 x=0 y=0 Z 1 Z x =− x=0 y=0 3x dy dx = − Z 1 x=0 ¡ ¢¯1 3x(x − 0) dx = − x3 ¯0 = −1. Problem 3.52 Verify Stokes’s theorem for the vector field B = (ˆrr cos φ + φˆ sin φ ) by evaluating: Z (a) B · dl over the semicircular contour shown in Fig. P3.52(a), and n ZC (b) S × B) · ds over the surface of the semicircle. (∇× y 2 y 2 L2 1 -2 L3 0 L1 (a) x 2 0 L2 L3 L4 L1 1 2 x (b) Figure P3.52: Contour paths for (a) Problem 3.52 and (b) Problem 3.53. Solution: (a) Z B · dl = n Z L1 B · dl + Z L2 B · dl + Z L3 B · dl, B · dl = (ˆrr cos φ + φˆ sin φ ) · (ˆr dr + φˆ r d φ + zˆ dz) = r cos φ dr + r sin φ d φ , µZ 2 µZ 0 ¶¯ ¶¯ ¯ ¯ r cos φ dr ¯¯ B · dl = + r sin φ d φ ¯¯ r=0 L1 φ =0 φ =0, z=0 z=0 ¡ 1 2 ¢¯2 = 2 r ¯r=0 + 0 = 2, ¶¯ ¶¯ µZ π µZ 2 Z ¯ ¯ r sin φ d φ ¯¯ B · dl = r cos φ dr ¯¯ + L φ =0 r=2 Z 2 Z L3 Z = 0+ µZ B · dl = z=0 π (−2 cos φ )|φ =0 = 4, ¶¯ 0 ¯ + r cos φ dr ¯¯ r=2 φ =π ,z=0 ¡ ¢¯0 = − 21 r2 ¯r=2 + 0 = 2, B · dl = 2 + 4 + 2 = 8. n r=2, z=0 π ¶¯ ¯ r sin φ d φ ¯¯ φ =π µZ z=0 (b) ∇×B = ∇×(ˆrr cos φ + φˆ sin φ ) µ ¶ µ ¶ 1 ∂ ∂ ∂ ∂ ˆ = rˆ 0 − (sin φ ) + φ (r cos φ ) − 0 r ∂φ ∂z ∂z ∂r ¶ µ ∂ 1 ∂ (r cos φ ) (r(sin φ )) − + zˆ r ∂r ∂φ ¶ µ 1 1 ˆ = rˆ 0 + φ 0 + zˆ (sin φ + (r sin φ )) = zˆ sin φ 1 + , r r µ ¶¶ ZZ Z π Z 2 µ 1 ∇×B · ds = · (ˆzr dr d φ ) zˆ sin φ 1 + r φ =0 r=0 Z π Z 2 ³¡ ¢¯2 ´¯¯π sin φ (r + 1) dr d φ = − cos φ ( 12 r2 + r) ¯r=0 ¯ = φ =0 r=0 φ =0 = 8. Problem 3.55 Verify Stokes’s theorem for the vector field B = (ˆr cos φ + φˆ sin φ ) by evaluating: (a) Z B · dℓℓ over the path comprising a quarter section of a circle, as shown in n C Fig. P3.55, and (b) Z S × B) · ds over the surface of the quarter section. (∇× y (0, 3) L1 x (−3, 0) L3 Figure P3.55: Problem 3.55. Solution: (a) Z B · dℓℓ = n C Z L1 B · dℓℓ + Z L2 B · dℓℓ + Z L3 B · dℓℓ Given the shape of the path, it is best to use cylindrical coordinates. B is already expressed in cylindrical coordinates, and we need to choose dℓℓ in cylindrical coordinates: dℓℓ = rˆ dr + φˆ r d φ + zˆ dz. Along path L1 , d φ = 0 and dz = 0. Hence, dℓℓ = rˆ dr and ¯ Z r=3 Z ¯ ˆ φ ℓ ˆ (ˆr cos φ + sin φ ) · r dr¯¯ B · dℓ = φ =90◦ r=0 L1 ¯ ¯ cos φ dr¯¯ = r=0 Z 3 φ =90◦ = Along L2 , dr = dz = 0. Hence, dℓℓ = φˆ r d φ and Z L2 B · dℓℓ = = Z 180◦ φ =90◦ ¯ ¯ r cos φ |3r=0 ¯ φ =90◦ ¯ ¯ (ˆr cos φ + φˆ sin φ ) · φˆ r d φ ¯ ◦ −3 cos φ |180 90◦ = 0. r=3 = 3. Along L3 , dz = 0 and d φ = 0. Hence, dℓℓ = rˆ dr and Z L3 B · dℓℓ = = Z 0 r=3 Z 0 r=3 Hence, Z ¯ ¯ (ˆr cos φ + φˆ sin φ ) · rˆ dr¯ φ =180◦ cos φ dr|φ =180◦ = −r|03 = 3. B · dℓℓ = 0 + 3 + 3 = 6. n C (b) µ µ ¶¶ 1 ∂ ∂ Br ∇ × B = zˆ rBφ − r ∂r ∂φ ¶ µ ∂ 1 ∂ (cos φ ) (r sin φ ) − = zˆ r ∂r ∂φ 2 1 = zˆ (sin φ + sin φ ) = zˆ sin φ . r r ¶ Z 3 Z 180◦ µ Z 2 (∇ × B) · ds = zˆ sin φ · zˆ r dr d φ r r=0 φ =90◦ S ¯180◦ ¯ = 6. = −2r|3r=0 cos φ ¯ ◦ φ =90 Hence, Stokes’s theorem is verified. Problem 3.57 Find the Laplacian of the following scalar functions: (a) V = 4xy2 z3 , (b) V = xy + yz + zx, (c) V = 3/(x2 + y2 ), (d) V = 5e−r cos φ , (e) V = 10e−R sin θ . Solution: (a) From Eq. (3.110), ∇2 (4xy2 z3 ) = 8xz3 + 24xy2 z. (b) ∇2 (xy + yz + zx) = 0. (c) From the inside back cover of the book, ¶ µ 12 3 2 = ∇2 (3r−2 ) = 12r−4 = . ∇ 2 2 x +y (x2 + y2 )2 (d) ∇ (5e 2 −r cos φ ) = 5e −r (e) ∇ (10e 2 −R sin θ ) = 10e −R ¶ µ 1 1 cos φ 1 − − 2 . r r µ ¸ · ¶ 2 cos2 θ − sin2 θ . sin θ 1 − + R R2 sin θ ...
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  • Fall '17
  • Daniele Tosi
  • Trigraph, Vector field

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