parametrics.ms (4516B)
1 .TL 2 Parametric equations 3 .AU 4 Lucas Standen 5 .AI 6 QMC 7 8 .EQ 9 delim @@ 10 .EN 11 .EQ 12 delim @# 13 .EN 14 15 .2C 16 17 .NH 1 18 What are they? 19 20 .LP 21 Parametric equations are two equations that are linked by a common variable usually @ t # 22 23 They are written in the format 24 .EQ 25 x = at 26 .EN 27 .EQ 28 y = bt 29 .EN 30 The can also include trig functions, exponents and other parts of maths 31 32 You may be asked to, convert to Cartesian, find the range and domain, differentiate and finding points of intersection 33 34 .NH 1 35 Converting to Cartesian 36 .LP 37 38 You will often be asked to convert to a Cartesian equivalent equation. To do that you will need to rearrange one of the given equations to get it in terms of @ t #, then substitute that value into the other equation. You will most often find that rearranging @ x # is easier as this will result in an equation equal to @ y #. 39 40 .NH 2 41 Example 42 .EQ 43 x = 2t 44 .EN 45 46 .EQ 47 y = t sup 2 48 .EN 49 50 .EQ 51 t = x over 2 52 .EN 53 54 .EQ 55 y = ({x over 2}) sup 2 56 .EN 57 58 .EQ 59 y = x sup 2 over 4 60 .EN 61 62 .NH 1 63 Finding the domain and range 64 .LP 65 The domain of the Cartesian equation is the range of the @ x # and the range of the Cartesian is the range of the @ y # equation. 66 67 .NH 2 68 Examples 69 .LP 70 .EQ 71 x = t - 2 72 .EN 73 74 .EQ 75 y = t sup 2 + 1 76 .EN 77 78 .EQ 79 "where" -4 <= t <= 4 80 .EN 81 82 .EQ 83 t = x + 2 84 .EN 85 86 .EQ 87 y = (x + 2) sup 2 + 1 88 .EN 89 90 .EQ 91 y = x sup 2 + 4x + 4 + 1 92 .EN 93 94 .EQ 95 y = x sup 2 + 4x + 5 96 .EN 97 98 .EQ 99 f(x) = x sup 2 + 4x + 5 100 .EN 101 102 .EQ 103 domain = -6 <= x <= 2 104 .EN 105 106 .EQ 107 range = 1 <= f(x) <= 16 108 .EN 109 110 .NH 1 111 Differentiating Parametric equations 112 .LP 113 This process is relatively simple, and can be solved by viewing @ dy over dx # as fractions. As we can already find @ dx over dt # and @ dy over dt # we can say the following. 114 115 .EQ 116 dy over dx = {dy over dt} over {dx over dt} 117 .EN 118 119 This is because the @ dt # will cancel out on the top and bottom. 120 121 .NH 2 122 Examples 123 .LP 124 125 .EQ 126 x = 2t 127 .EN 128 129 .EQ 130 y = t sup 2 - 3t + 2 131 .EN 132 133 .EQ 134 dx over dt = 2 135 .EN 136 137 .EQ 138 dy over dt = 2t - 3 139 .EN 140 141 .EQ 142 {dy over dt} over {dx over dt} = 2 over {2t -3} 143 .EN 144 145 .NH 1 146 Points of intersection 147 .LP 148 This is as simple as substituting in numbers for the most part although sometimes it can be harder due to trig identities showing up. 149 150 .NH 2 151 Example 152 .LP 153 Lets say we have the parametric @ x = t sup 2 # and @ y = 8t - 5 # and we want to know if there is any points of 154 intersection with the line @ y = 5x + 4 # 155 156 To solve this we can find the Cartesian equation and then do this normally. 157 .EQ 158 t = sqrt x 159 .EN 160 161 .EQ 162 y = 8{sqrt x} - 5 163 .EN 164 165 .EQ 166 "Then we can say" 167 .EN 168 169 .EQ 170 5x + 4 = 8{sqrt x} - 5 171 .EN 172 173 .EQ 174 5x + 9 = 8{sqrt x} 175 .EN 176 177 .EQ 178 5x over 8 + 9 over 8 = sqrt x 179 .EN 180 181 .EQ 182 ({5x over 8 + 9 over 8}) sup 2 = x 183 .EN 184 185 .EQ 186 {25x sup 2 over 64 + 81 over 64} = x 187 .EN 188 189 .EQ 190 25x sup 2 + 81 = 64x 191 .EN 192 193 .EQ 194 25x sup 2 -64x + 81 = 0 195 .EN 196 197 .EQ 198 "This has no points of intersection because x is imaginary" 199 .EN 200 201 .LP 202 EX 8D Q1a 203 .EQ 204 x = 5 + t 205 .EN 206 .EQ 207 y = 6 - t 208 .EN 209 .EQ 210 y = 6 - (x - 5) 211 .EN 212 .EQ 213 y = 11 - x 214 .EN 215 .EQ 216 0 = 11 - x 217 .EN 218 .EQ 219 -11 = - x 220 .EN 221 222 .EQ 223 11 = x 224 .EN 225 .LP 226 227 EX 8D 12b 228 .EQ 229 x = 6cos(t) 230 .EN 231 .EQ 232 y = 4sin(2t) + 2 233 .EN 234 .EQ 235 -{pi over 2} < t < {pi over 2} 236 .EN 237 238 .EQ 239 4 = 4sin(2t) + 2 240 .EN 241 242 .EQ 243 2 = 4sin(2t) 244 .EN 245 246 .EQ 247 1 over 2 = sin(2t) 248 .EN 249 250 .EQ 251 2t = arcsin({1 over 2}) 252 .EN 253 254 .EQ 255 2t = pi over 6, {5 pi} over 6 256 .EN 257 258 .EQ 259 t = pi over 12, {5 pi} over 12 260 .EN 261 262 .LP 263 EX 8D 12c 264 265 .EQ 266 x = 6cos(t) 267 .EN 268 269 .EQ 270 x = 3 sqrt 3, - 3 sqrt 3 271 .EN 272 273 .EQ 274 (3 sqrt 3, 4), (- 3 sqrt 3, 4) 275 .EN 276 277 .LP 278 EX 8D 13 279 280 .EQ 281 x = 2t 282 .EN 283 .EQ 284 y = 4t sup 2 - 4t 285 .EN 286 287 .EQ 288 t = x over 2 289 .EN 290 291 .EQ 292 y = 4({x over 2}) sup 2 - 4 ({x over 2}) 293 .EN 294 295 .EQ 296 y = 4({x sup 2 over 4}) - 4 ({x over 2}) 297 .EN 298 299 .EQ 300 y = x sup 2 - 4 ({x over 2}) 301 .EN 302 303 .EQ 304 y = x sup 2 - 2x 305 .EN 306 307 .EQ 308 2x - 5 = x sup 2 - 2x 309 .EN 310 311 .EQ 312 0 = x sup 2 - 4x + 5 313 .EN 314 315 .EQ 316 sqrt {-4 sup 2 - 4 (1) (5) } 317 .EN 318 319 .EQ 320 sqrt {16 - 20 } 321 .EN 322 323 .EQ 324 sqrt {-4} 325 .EN 326 327 .EQ 328 sqrt {-4} " has no real solutions!" 329 .EN 330 331 .NH 1 332 Parametric with trig 333 .LP 334 Trig often shows up in parametric equations, however using the identities we know, they can be easy to solve. The reason they are harder, is because we can't rearrange to get @ t # like in other questions. 335 336 What we have to do is to use the identities we know, such as 337 .EQ 338 sin sup 2 (x) + cos sup 2 (x) = 1 339 .EN 340 341 An example would be the following 342 343 .EQ 344 x = 2sin(t) 345 .EN 346 347 .EQ 348 y = cos(t) + 2 349 .EN 350 351 .EQ 352 sin(t) = x over 2 353 .EN 354 355 .EQ 356 cos(t) = y - 2 357 .EN 358 359 .EQ 360 ({x over 2}) sup 2 + (y - 2) sup 2 = 1 361 .EN 362 363 .EQ 364 {x sup 2 over 4} + y sup 2 -4y + 4 = 1 365 .EN 366 367 .EQ 368 {x sup 2 over 4} + y sup 2 -4y + 4 = 1 369 .EN 370 371 .EQ 372 {x sup 2} + 4y sup 2 -16y + 16 = 4 373 .EN 374 375 .EQ 376 {x sup 2} -16y + 16 - 4= -4y sup 2 377 .EN 378 379 .EQ 380 {x sup 2} + 12 = -4y sup 2 + 16y 381 .EN 382 383 This is an oval shape.