main.typ (13102B)
1 #let tag(equation, tag) = math.equation(numbering: _ => tag, block: true, equation) 2 #let vec(v) = math.bold(math.upright(v)) 3 #let unit(u) = math.hat(math.bold(math.upright(u))) 4 #let cross = math.times 5 #let definedAs = math.eq.triple 6 #let implies = math.arrow.r.double 7 #let sepMag = text(font: "Script R", "r") 8 #let sepVec = text(font: "Script R", weight: "bold", "r") 9 #let sepUnit = math.hat(sepVec) 10 11 #set heading(numbering: (..nums) => { 12 let n = nums.pos() 13 if n.len() <= 2 { 14 numbering("1.1", ..n) 15 } 16 }) 17 18 = Vector Analysis 19 20 == Vector Algebra 21 22 === Problem 1.1 23 24 === Problem 1.2 25 26 The cross product is not associative. We will show this by providing a counterexample. 27 28 Let $vec(A)$, $vec(B)$, and $vec(C)$ be vectors, with $vec(C)$ perpendicular to both $vec(A)$ and 29 $vec(B)$, and $vec(A)$ and $vec(B)$ neither perpendicular nor parallel. 30 31 $vec(A) cross vec(B)$ is equal to $A B sin theta_(vec(A) angle vec(B)) unit(n)_(vec(A) cross 32 vec(B))$ by definition. Since $vec(C)$ is perpendicular to both $vec(A)$ and $vec(B)$, their cross 33 product is parallel to $vec(C)$. The cross product of any two parallel vectors is $vec(0)$ because 34 the angle between them is $0$, and $sin (0) = 0$. So, $(vec(A) cross vec(B)) cross vec(C) = vec(0)$. 35 36 $vec(B) cross vec(C)$ is equal to $B C sin theta_(vec(B) angle vec(C)) unit(n)_(vec(B) cross 37 vec(C))$ by definition. The result is perpendicular to $vec(C)$, which means that it must be in the 38 plane of $vec(A)$ and $vec(B)$. But, the result is not parallel to $vec(A)$ because $vec(B)$ is not 39 perpendicular to $vec(A)$. So, $vec(A) cross (vec(B) cross vec(C)) eq.not vec(0)$. 40 41 The left and right sides of the definition of associativity are not equal for this set of vectors, 42 which means they are a counterexample. Therefore the cross product is not associative. $qed$ 43 44 === Problem 1.3 45 46 We will use the same cube arrangement as Example 1.2. Let $vec(A)$ be the body diagonal extending 47 from the origin to $(1, 1, 1)$, and let $vec(B)$ be the body diagonal extending from $(1, 0, 0)$ to 48 $(0, 1, 1)$. In component form, these are 49 50 $ 51 vec(A) = (1 - 0) unit(x) + (1 - 0) unit(y) + (1 - 0) unit(z) = unit(x) + unit(y) + unit(z) \ 52 vec(B) = (0 - 1) unit(x) + (1 - 0) unit(y) + (1 - 0) unit(z) = -unit(x) + unit(y) + unit(z) "." 53 $ 54 55 The magnitudes of these vectors are both $sqrt(3)$. 56 57 $ 58 A = sqrt(vec(A) dot vec(A)) = sqrt(1 dot 1 + 1 dot 1 + 1 dot 1) = sqrt(3) \ 59 B = sqrt(vec(B) dot vec(B)) = sqrt((-1) dot (-1) + 1 dot 1 + 1 dot 1) = sqrt(3) 60 $ 61 62 We can find the angle between these vectors by first computing the dot product in component form and 63 then isolating the angle $theta$ in abstract form. 64 65 $ 66 vec(A) dot vec(B) = 1 dot (-1) + 1 dot 1 + 1 dot 1 = 1 \ 67 vec(A) dot vec(B) definedAs A B cos theta_(vec(A) vec(B)) implies theta_(vec(A) vec(B)) = cos^(-1) ((vec(A) dot vec(B)) / (A B)) \ 68 theta_(vec(A) vec(B)) = cos^(-1) (1 / (sqrt(3) sqrt(3))) = cos^(-1) (1 / 3) approx #rect($1.231 "radians"$) 69 $ 70 71 Note that the dot product implicitly considers vectors that all extend from the same origin, so by 72 applying it to $vec(B)$ we have "moved" it so that its tail is at the same origin as $vec(A)$. This 73 is acceptable because it does not change the magnitude or direction of the vector. 74 75 === Problem 1.4 76 77 The cross product of any two non-parallel vectors points normal to the plane formed by the two 78 inputs. For convenience, we select two of the edges of the shaded triangle to be our input vectors 79 such that their cross product will have the same direction as $unit(n)$. 80 81 Let $vec(A)$ be the edge that lies in the $x y$ plane, and let $vec(B)$ be the edge extending from 82 $(1, 0, 0)$ to $(0, 0, 3)$. In component form, these are 83 84 $ 85 vec(A) = (0 - 1) unit(x) + (2 - 0) unit(y) + (0 - 0) unit(z) = -unit(x) + 2 unit(y) \ 86 vec(B) = (0 - 1) unit(x) + (0 - 0) unit(y) + (3 - 0) unit(z) = -unit(x) + 3 unit(z) "." 87 $ 88 89 Their cross product is 90 91 $ 92 vec(A) cross vec(B) &= (A_y B_z - A_z B_y) unit(x) + (A_z B_x - A_x B_z) unit(y) + (A_x B_y - A_y B_x) unit(z) \ 93 &= (2 dot 3 - 0 dot 0) unit(x) + (0 dot (-1) - (-1) dot 3) unit(y) + ((-1) dot 0 - 2 dot (-1)) unit(z) \ 94 &= 6 unit(x) + 3 unit(y) + 2 unit(z) "." 95 $ 96 97 The result of $vec(A) cross vec(B)$ has the same direction as $unit(n)$, but it is not a unit 98 vector. We must divide by its magnitude to compute $unit(n)$. 99 100 $ 101 unit(n) = (vec(A) cross vec(B)) / (|vec(A) cross vec(B)|) 102 = (6 unit(x) + 3 unit(y) + 2 unit(z)) / (sqrt(6^2 + 3^2 + 2^2)) 103 = 1/7 (6 unit(x) + 3 unit(y) + 2 unit(z)) 104 $ 105 106 The components of $unit(n)$ are $#rect($unit(n)_x = 6/7, unit(n)_y = 3/7, "and" unit(n)_z = 2/7 107 "."$)$ 108 109 === Problem 1.5 110 111 The *BAC-CAB* rule is 112 113 #tag($ vec(A) cross (vec(B) cross vec(C)) = vec(B) (vec(A) dot vec(C)) - vec(C) (vec(A) dot vec(B)) $, "(Eq. 1.17)") 114 115 which we can prove by expanding both sides in component form. We will evaluate each side in 116 isolation and then compare them. 117 118 First, we evaluate the left side. 119 120 $ 121 vec(A) cross (vec(B) cross vec(C)) =& 122 vec(A) cross ((B_y C_z - B_z C_y) unit(x) 123 + (B_z C_x - B_x C_z) unit(y) 124 + (B_x C_y - B_y C_x) unit(z)) \ 125 =& (A_y (B_x C_y - B_y C_x) - A_z (B_z C_x - B_x C_z)) unit(x) \ 126 &+ (A_z (B_y C_z - B_z C_y) - A_x (B_x C_y - B_y C_x)) unit(y) \ 127 &+ (A_x (B_z C_x - B_x C_z) - A_y (B_y C_z - B_z C_y)) unit(z) \ 128 =& (A_y (B_x C_y - B_y C_x) - A_z (B_z C_x - B_x C_z)) unit(x) \ 129 &+ (A_z (B_y C_z - B_z C_y) - A_x (B_x C_y - B_y C_x)) unit(y) \ 130 &+ (A_x (B_z C_x - B_x C_z) - A_y (B_y C_z - B_z C_y)) unit(z) \ 131 =& (A_y B_x C_y + A_z B_x C_z - A_y B_y C_x - A_z B_z C_x) unit(x) \ 132 &+ (A_x B_y C_x + A_z B_y C_z - A_x B_x C_y - A_z B_z C_y) unit(y) \ 133 &+ (A_x B_z C_x + A_y B_z C_y - A_x B_x C_z - A_y B_y C_z) unit(z) 134 $ 135 136 And now the right side. 137 138 $ 139 vec(B) (vec(A) dot vec(C)) - vec(C) (vec(A) dot vec(B)) =& 140 vec(B) (A_x C_x + A_y C_y + A_z C_z) - vec(C) (A_x B_x + A_y B_y + A_z B_z) \ 141 =& B_x (A_x C_x + A_y C_y + A_z C_z) unit(x) \ 142 &+ B_y (A_x C_x + A_y C_y + A_z C_z) unit(y) \ 143 &+ B_z (A_x C_x + A_y C_y + A_z C_z) unit(z) \ 144 &- C_x (A_x B_x + A_y B_y + A_z B_z) unit(x) \ 145 &- C_y (A_x B_x + A_y B_y + A_z B_z) unit(y) \ 146 &- C_z (A_x B_x + A_y B_y + A_z B_z) unit(z) \ 147 =& (A_x B_x C_x + A_y B_x C_y + A_z B_x C_z - A_x B_x C_x - A_y B_y C_x - A_z B_z C_x) unit(x) \ 148 &+ (A_x B_y C_x + A_y B_y C_y + A_z B_y C_z - A_x B_x C_y - A_y B_y C_y - A_z B_z C_y) unit(y) \ 149 &+ (A_x B_z C_x + A_y B_z C_y + A_z B_z C_z - A_x B_x C_z - A_y B_y C_z - A_z B_z C_z) unit(z) \ 150 =& (A_y B_x C_y + A_z B_x C_z - A_y B_y C_x - A_z B_z C_x) unit(x) \ 151 &+ (A_x B_y C_x + A_z B_y C_z - A_x B_x C_y - A_z B_z C_y) unit(y) \ 152 &+ (A_x B_z C_x + A_y B_z C_y - A_x B_x C_z - A_y B_y C_z) unit(z) 153 $ 154 155 Both sides are equal, which proves the *BAC-CAB* rule. $qed$ 156 157 === Problem 1.6 158 159 === Problem 1.7 160 161 The separation vector $sepVec$ is defined as 162 163 #tag($ sepVec definedAs vec(r) - vec(r') "." $, "(Eq 1.23)") 164 165 The field vector is $vec(r) = 4 unit(x) + 6 unit(y) + 8 unit(z)$, and the source vector is $vec(r') 166 = 2 unit(x) + 8 unit(y) + 7 unit(z)$, so the separation vector is 167 168 $ 169 sepVec = (4 - 2) unit(x) + (6 - 8) unit(y) + (8 - 7) unit(z) \ 170 #rect($sepVec = 2 unit(x) - 2 unit(y) + unit(z)$) "." 171 $ 172 173 The magnitude of the separation vector is 174 175 $ 176 sepMag = sqrt(sepMag_x^2 + sepMag_y^2 + sepMag_z^2) = sqrt(2^2 + (-2)^2 + 1^2) = sqrt(9) \ 177 #rect($sepMag = 3$) "." 178 $ 179 180 The unit vector associated with the separation vector is 181 182 $ 183 sepUnit = sepVec / sepMag = (2 unit(x) - 2 unit(y) - unit(z)) / 3 \ 184 #rect($sepUnit = 2/3 unit(x) - 2/3 unit(y) + 1/3 unit(z)$) "." 185 $ 186 187 === Problem 1.8 188 189 === Problem 1.9 190 191 === Problem 1.10 192 193 == Differential Calculus 194 195 === Problem 1.11 196 197 #tag($ 198 gradient T definedAs (partial T) / (partial x) unit(x) 199 + (partial T) / (partial y) unit(y) 200 + (partial T) / (partial z) unit(z) 201 $, "(Eq. 1.36)") 202 203 ==== Part (a) 204 205 $ 206 gradient f = gradient (x^2 + y^3 + z^4) \ 207 gradient f = partial / (partial x) (x^2 + y^3 + z^4) unit(x) 208 + partial / (partial y) (x^2 + y^3 + z^4) unit(y) 209 + partial / (partial z) (x^2 + y^3 + z^4) unit(z) \ 210 #rect($ gradient f = 2 x unit(x) + 3 y^2 unit(y) + 4 z^3 unit(z) $) 211 $ 212 213 ==== Part (b) 214 215 $ 216 gradient f = gradient (x^2 y^3 z^4) \ 217 gradient f = partial / (partial x) (x^2 y^3 z^4) unit(x) 218 + partial / (partial y) (x^2 y^3 z^4) unit(y) 219 + partial / (partial z) (x^2 y^3 z^4) unit(z) \ 220 #rect($ gradient f &= 2 x y^3 z^4 unit(x) + 3 x^2 y^2 z^4 unit(y) + 4 x^2 y^3 z^3 unit(z) $) 221 $ 222 223 ==== Part (c) 224 225 $ 226 gradient f = gradient (e^x sin(y) ln(z)) \ 227 gradient f = partial / (partial x) lr((e^x sin(y) ln(z)), size: #200%) unit(x) 228 + partial / (partial y) lr((e^x sin(y) ln(z)), size: #200%) unit(y) 229 + partial / (partial z) lr((e^x sin(y) ln(z)), size: #200%) unit(z) \ 230 #rect($ gradient f &= e^x unit(x) + cos(y) unit(y) + 1/z unit(z) $) 231 $ 232 233 === Problem 1.12 234 235 ==== Part (a) 236 237 The top of the hill is the maximum of $h$. The gradient at this point is zero, so we can find the 238 maximum by computing $gradient h$ and finding the values of $x$ and $y$ for which the resulting 239 vector is $vec(0)$. In the case of multiple points with zero gradient, we take the point which 240 maximizes $h$. 241 242 First, we compute the gradient of $h$. 243 244 $ 245 gradient h =& partial / (partial x) (10 (2 x y - 3 x^2 - 4 y^2 - 18 x + 28 y + 12)) unit(x) \ 246 +& partial / (partial y) (10 (2 x y - 3 x^2 - 4 y^2 - 18 x + 28 y + 12)) unit(y) \ 247 gradient h =& 20 (-3 x + y - 9) unit(x) + 20 (x - 4 y + 14) unit(y) 248 $ 249 250 Now we set $gradient h$ equal to $vec(0)$ and solve the resulting system of equations. 251 252 $ 253 20 (-3 x + y - 9) = 0 \ 254 20 ( x - 4 y + 14) = 0 \ 255 implies y = 4 x + 9 \ 256 x - 3 (4 x + 9) + 14 = 0 implies x = -13/12 \ 257 y = 3 (-13/12) + 9 = 5 3/4 258 $ 259 260 There is only one solution, and we assume that this point is a maximum instead of a minimum or 261 saddle point. 262 263 So the top of the hill is located $#rect($1 1/12 "miles west, " 5 3/4 "miles north"$)$ of South 264 Hadley. 265 266 ==== Part (b) 267 268 The value of $h$ at the top of the hill is is 269 270 $ 271 h(-13/8, 33/8) &= 10 (2 x y - 3 x^2 - 4 y^2 - 18 x + 28 y + 12) 272 &approx 44.3 273 $ 274 275 So the hill is $#rect($"about" 44.3 "feet"$)$ high. 276 277 ==== Part (c) 278 279 To find the slope and steepest direction at the given point, we evaluate the gradient there and find 280 its magnitude and direction. 281 282 $ 283 (gradient h)_(x=1,y=1) &= 20 (-3 dot 1 + 1 - 9) unit(x) + 20 (1 - 4 dot 1 + 14) unit(y) = -220 unit(x) + 220 unit(y) \ 284 abs((gradient h)_(x=1,y=1)) &= sqrt((-220)^2 + 220^2) approx 221 \ 285 $ 286 287 The gradient at the given point may be converted directly into a bearing relative to geographic 288 north. 289 290 $ 291 tan^(-1) ((-220)/220) + 2 pi approx 5.50 approx 315 degree 292 $ 293 294 So the steepest slope at the given point is $#rect($"about" 221 "feet per mile at a bearing of" 315 295 degree "."$)$ 296 297 For completeness, we also compute the unit vector of the gradient at the given point. 298 299 $ 300 (gradient h)/abs(gradient h)|_(x=1,y=1) = (-220 unit(x) + 220 unit(y))/(sqrt((-220)^2 + 220^2)) 301 approx -sqrt(2)/2 unit(x) + sqrt(2)/2 unit(y) 302 $ 303 304 === Problem 1.13 305 306 $ 307 sepVec = (x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z) \ 308 sepMag = sqrt((x - x')^2 + (y - y')^2 + (z - z')^2) 309 $ 310 311 ==== Part (a) 312 313 $ 314 gradient (sepMag^2) &= 2 sepVec \ 315 partial / (partial x) sepMag^2 unit(x) + partial / (partial y) sepMag^2 unit(y) + partial / (partial z) sepMag^2 unit(z) 316 &= 2 sepVec \ 317 2 (x - x') unit(x) + 2 (y - y') unit(y) + 2 (z - z') unit(z) &= 2 sepVec \ 318 2 sepVec &= 2 sepVec && qed 319 $ 320 321 ==== Part (b) 322 323 $ 324 gradient (1/sepMag) = -sepUnit/sepMag^2 \ 325 partial / (partial x) 1/sepMag unit(x) + partial / (partial y) 1/sepMag unit(y) + partial / (partial z) 1/sepMag unit(z) 326 = -((x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z))/(sqrt((x - x')^2 + (y - y')^2 + (z - z')^2)) 327 1/((x - x')^2 + (y - y')^2 + (z - z')^2) \ 328 -(2 (x - x'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) unit(x) \ 329 -(2 (y - y'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) unit(y) \ 330 -(2 (z - z'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) unit(z) 331 = -sepVec / sepMag^3 \ 332 -((x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z))/(((x - x')^2 + (y - y')^2 + (z - z')^2)^(3/2)) = -sepVec / sepMag^(3/2) \ 333 -sepVec / sepMag^3 = -sepVec / sepMag^3 qed 334 $ 335 336 ==== Part (c) 337 338 $ 339 gradient sepMag^n =& partial / (partial x) sepMag^n unit(x) + partial / (partial y) sepMag^n unit(y) 340 + partial / (partial z) sepMag^n unit(z) \ 341 =& (n (x - x'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2)) unit(x) \ 342 &+ (n (y - y'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2)) unit(y) \ 343 &+ (n (z - z'))/(2 ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2)) unit(z) \ 344 =& n/2 ((x - x') unit(x) + (y - y') unit(y) + (z - z') unit(z)) / ((x - x')^2 + (y - y')^2 + (z - z')^2)^(n/2) \ 345 $ 346 347 $ 348 #rect($ gradient sepMag^n = n/2 sepVec / sepMag^(n) $) 349 $ 350 351 === Problem 1.14 352 353 === Problem 1.15 354 355 === Problem 1.16 356 357 === Problem 1.17 358 359 === Problem 1.18 360 361 === Problem 1.19 362 363 === Problem 1.20 364 365 === Problem 1.21 366 367 === Problem 1.22 368 369 === Problem 1.23 370 371 === Problem 1.24 372 373 === Problem 1.25 374 375 === Problem 1.26 376 377 === Problem 1.27 378 379 === Problem 1.28 380 381 == Integral Calculus 382 383 == Curvilinear Coordinates 384 385 == The Dirac Delta Function 386 387 == The Theory of Vector Fields 388 389 = Electrostatics 390 391 = Potentials 392 393 = Electric Fields in Matter 394 395 = Magnetostatics 396 397 = Magnetic Fields in Matter 398 399 = Electrodynamics 400 401 = Conservation Laws 402 403 = Electromagnetic Waves 404 405 = Potentials and Fields 406 407 = Radiation 408 409 = Electrodynamics and Relativity