Mt 2C Multivrible Clculus Lecture 5 1 Lines nd plnes Slide 1 Equtions of lines (Vector, prmetric, nd symmetric eqs.). Equtions of plnes. Distnce from point to plne. Equtions of lines Slide 2 Definition 1 Let P be point in spce, nd v be nonzero vector. Fix coordinte system wit origin t O, nd let r = OP. Ten, te set of vectors r(t) = r + tv, t IR, is clled te line troug P prllel to v. Tis is te vector eqution of te line.
Mt 2C Multivrible Clculus Lecture 5 2 Equtions of lines Consider te cse of 3 dimensions. In components, Slide 3 ten one s r(t) = x(t), y(t), z(t), r = x, y, z, v = v x, v y, v z, x(t) = x + tv x, y(t) = y + tv y, z(t) = z + tv z. Tese re clled te prmetric equtions of te line. Equtions of lines Compute t in expressions bove, nd denote x = x(t), y = y(t), nd z = z(t). Ten, t = x x v x = y y v y = z z v z. Slide 4 Tese re clled te symmetric equtions of te line. Definition 2 Two lines r(t) = r + tv, nd r(t) = r + tṽ re prllel if nd only if v = ṽ, wit. Notice tt in 2 dimensions, two lines re eiter prllel or tey intersect (or bot, wen tey coincide). Tis is not true in 3 dimensions. Two lines in 3 dimensions re clled skew lines if tey re neiter prllel nor tey intersect.
Mt 2C Multivrible Clculus Lecture 5 3 Equtions of plnes Definition 3 Fix point in spce, P, nd nonzero vector n. Te set of ll points P stisfying P P n = Slide 5 is clled te plne pssing troug P norml to n, nd we denote it s (P, n). Teorem 1 Fix coordinte system wit origin t O. Te eqution of te plne pssing troug P norml to n cn be written s (r r ) n =, were r = OP, nd r = OP, wit P in te plne. (Proof: P P = r r.) Equtions of plnes Slide 6 Teorem 2 In components, te eqution of plne (P, n) s te form were (x x )n x + (y y )n y + (z z )n z =, r = x, y, z, n = n x, n y, n z, r = x, y, z. Definition 4 Two plnes re prllel if teir normls re prllel. Teorem 3 Te ngle between two non-prllel plnes is te ngle between teir norml vectors.
Mt 2C Multivrible Clculus Lecture 6 4 Distnce from point to plne Slide 7 Teorem 4 Te distnce d from point P 1 to plne (P, n) is te sortest distnce from P 1 to ny point in te plne, nd is given by te expression d = P P n. (1) n (Proof: Drw picture of te sitution. Ten, were θ is te ngle between d = P P cos(θ), P P nd n. Reclling tt P P n = n P P cos(θ), nd tt te distnce is nonnegtive number, one gets Eq. (1).) Vector vlued functions nd derivtives Slide 8 Definition of Vector vlued functions. Limits Derivtive of vector vlued functions. Definition. Derivtive rules. Definite integrls.
Mt 2C Multivrible Clculus Lecture 6 5 Vector vlued functions Definition 5 A vector vlued function r(t) is function r : D IR R IR n, Slide 9 wit n 2. Te symbol mens subset of. Te set D is clled domin of r, nd R is te rnge of r. In components, n = 3, r(t) = x(t), y(t), z(t). Tink vector vlued function s 3 usul functions. (Usul mens sclr vlued functions f : IR IR.) Vector vlued functions Slide 1 Tere is nturl ssocition between curves in IR n nd vector vlued functions. Te curve is determined by te ed points of te vector vlued function r(t). Te independent vrible t is clled te prmeter of te curve.
Mt 2C Multivrible Clculus Lecture 6 6 Limits Slide 11 Definition 6 Consider te function r(t) = x(t), y(t), z(t). Te limit lim t t r(t) is defined s lim r(t) = lim x(t), lim y(t), lim z(t), t t t t t t t t wen suc limit in ec component exists. Terefore, our definition is: Te limit of vector vlued functions s t pproces t is te limit of its components in Crtesin coordinte system. Definition 7 A function r(t) is continuous t t if lim t t r(t) = r(t ). Hving te ide of limit, one cn introduce te ide of derivtive of vector vlued function. Derivtive Definition 8 Te derivtive of vector vlued function r : IR IR n, denoted r (t), is given by Slide 12 wen suc limit exists. r r(t + ) r(t) (t) = lim, Note: Te derivtive of vector vlued function is vector. Drwing n pproprite picture one concludes tt r (t) is vector tngent to te curve given by r(t). If r(t) represents te vector position of prticle, ten r (t) represents te velocity vector of tt prticle. Tt is, v(t) = r (t).
Mt 2C Multivrible Clculus Lecture 6 7 Teorem 5 Consider te function r(t) = x(t), y(t), z(t), were x(t), y(t), nd z(t) re differentible functions. Ten r (t) = x (t), y (t), z (t). Slide 13 (Proof: = r (t) = lim r(t+) r(t), = lim x(t+) x(t) lim x(t+) x(t), y(t+) y(t), lim y(t+) y(t), z(t+) z(t), lim z(t+) z(t) = x (t), y (t), z (t). ) Differentition rules [v(t) + w(t)] = v (t) + w (t), (ddition); [cv(t)] = cv (t), (product rule for constnts); Slide 14 [f(t)v(t)] = f (t)v(t) + f(t)v (t), (product rule for sclr functions); [v(t) w(t)] = v (t) w(t) + v(t) w (t), (product rule for dot product); [v(t) w(t)] = v (t) w(t) + v(t) w (t), (product rule for cross product); [v(f(t))] = v (f(t))f (t), (cin rule for functions).
Mt 2C Multivrible Clculus Lecture 6 8 Higer derivtives Te m derivtive of r(t) is denoted s r (m) (t) nd is given by te expression r (m) (t) = [r (m 1) (t)]. Exmple: Slide 15 r(t) = cos(t), sin(t), t 2 + 2t + 1, r (t) = sin(t), cos(t), 2t + 2, r (2) (t) = (r (t)) = cos(t), sin(t), 2, r (3) (t) = (r (2) (t)) = sin(t), cos(t),. If r(t) is te vector position of prticle, ten te velocity vector is v(t) = r (t), nd te ccelertion vector is (t) = v (t) = r (2) (t). Definite integrls Slide 16 Definition 9 Te definite integrl of r(t) form t [, b] is vector wose components re te integrls of te components of r(t), nmely, b b b b r(t)dt = x(t)dt, y(t)dt, z(t)dt. Exmple: π cos(t), sin(t), t dt = = = π cos(t)dt, π sin(t)dt, sin(t) π, cos(t) π, t2 2, 2, π2. 2 π π,, tdt,
Mt 2C Multivrible Clculus Lecture 7 9 Comment Te definitions of limit nd derivtive of vector vlued functions were introduced component by component in fixed, given in dvnce, Crtesin coordinte system. Slide 17 Wt does ppen if one needs to work in oter coordinte system, sy, spericl coordintes, or rbitrry coordintes? All te notions of limit nd derivtives for vector vlued functions cn be generlized in wy tt te Crtesin coordintes system need not to be introduced. Wen one introduce suc Crtesin coordintes systems, one recovers te definitions presented ere. We do not study in our course tis coordinte independent notion of limit nd derivtives. Slide 18 Arc lengt nd rc lengt function Arc lengt of curve. Arc lengt function.
Mt 2C Multivrible Clculus Lecture 7 1 Arc lengt of curve Te rc lengt of curve in spce is number. It mesures te extension of te curve. Definition 1 Te rc lengt of te curve ssocited to vector vlued function r(t), for t [, b] is te number given by Slide 19 l b = b r (t) dt. Suppose tt te curve represents te pt trveled by prticle in spce. Ten, te definition bove sys tt te lengt of te curve is te integrl of te speed, v(t). So we sy tt te lengt of te curve is te distnce trveled by te prticle. Te formul bove cn be obtined s limit procedure, dding up te lengts of polygonl line tt pproximtes te originl curve. In components, one s, Arc lengt of curve Slide 2 r(t) = x(t), y(t), z(t), r (t) = x (t), y (t), z (t), r (t) = [x (t)] 2 + [y (t)] 2 + [z (t)] 2, b l b = [x (t)] 2 + [y (t)] 2 + [z (t)] 2 dt. Te rc lengt of generl curve could be very rd to compute.
Mt 2C Multivrible Clculus Lecture 7 11 Arc lengt function Definition 11 Consider vector vlued function r(t). Te rc lengt function l(t) from t = t is given by Slide 21 l(t) = t r (u) du. Note: l(t) is sclr function. It stisfies l(t ) =. Note: Te function l(t) represents te lengt up to t of te curve given by r(t). Our min ppliction: Reprmetriztion of given vector vlued function r(t) using te rc lengt function. Arc lengt function Reprmetriztion of curve using te rc lengt function: Wit r(t) compute l(t), strting t some t = t. Slide 22 Invert te function l(t) to find te function t(l). Exmple: l(t) = 3e t/2, ten t(l) = 2 ln(l/3). Compute te composition r(l) = r(t(l)). Tt is, replce t by t(l). Te function r(l) is te reprmetriztion of r(t) using te rc lengt s te new prmeter.