15. Vector Valued Functions


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1 1. Vecor Valued Funcions Up o his poin, we have presened vecors wih consan componens, for example, 1, and,,4. However, we can allow he componens of a vecor o be funcions of a common variable. For example, r() = + 1, + 3 presens a funcion whose inpu is a scalar, and whose oupu is a vecor in R. Such a funcion is called a vecorvalued funcion and is called a parameer variable. The common noaion is o wrie r() = x(), y() for vecorvalued funcions in R, and r() = x(), y(), z() for vecorvalued funcions in R 3. The number of parameer variables can be greaer han one. Example 1.1: Skech r() = + 1, + 3 for 1. Soluion: Le s build an inpuoupu able: r() = + 1, r( 1) = ( 1) + 1, ( 1) + 3 = 1,4. r(.) = (.) + 1, (.) + 3 =, 3. r() = () + 1, () + 3 = 1,3. r(.) = (.) + 1, (.) + 3 =, 3. 1 r(1) = (1) + 1, (1) + 3 = 3,4 1. r(1.) = (1.) + 1, (1.) + 3 = 4,. r() = () + 1, () + 3 =,7 We hen skech vecors for each such ha is foo is a he origin: This looks like a mess, bu i is a ruhful and lieral represenaion of r() = + 1, + 3 for cerain values of in he inerval 1. However, when represening he graph of a vecor valued funcion, i is common o only show he posiion a he head of he vecor, and he curve ha resuls. 76
2 This image is much cleaner and we see ha he pah raced ou by he heads of he vecors given by r() = + 1, + 3 for 1 form a parabola. Noe ha some of he values are saed a cerain poins. I is common o place an arrow on his pah o show he direcion of increasing value of he variable. Example 1.: Skech r() = a cos, a sin, for π, and describe he curve ha is raced ou by he vecors. Soluion: We build an inpuoupu able: r() = a cos, a sin r() = a cos, a sin = a, π 4 r(π 4) = a cos(π 4), a sin(π 4) = a, a π r(π ) = a cos(π ), a sin(π ) =, a 3π 4 r(3 π 4) = a cos(3 π 4), a sin(3π 4) = a, a π r(π) = a cos(π), a sin(π) = a, π 4 r(π 4) = a cos(π 4), a sin(π 4) = a, a 3π r(3π ) = a cos(3π ), a sin(3π ) =, a 7π 4 r(7π 4) = a cos(7π 4), a sin(7 π 4) = a, a π r(π) = a cos(π), a sin(π) = a, The curve is on he following page. The vecors are no acually drawn. Insead, he curve formed by he placemen of each vecor s head is drawn. 77
3 The curve is a circle of radius a, cenered a he origin. The bounds π ensure ha exacly one revoluion of he circle is skeched. Noe ha cerain poins on he pah are given by ordered pairs. Remember ha hese are he heads of he vecors, which are no drawn. Thus, he poin (, a) represens he head of he vecor, a when = π. The arrow shows he direcion of increasing, and he circle sars a he poin (a, ) and ends a his same poin, one revoluion laer. There is more han one way o define a circle of radius a. For example, r() = a sin, a cos, for π races he same circle, bu his ime saring a (, a) and in he clockwise direcion. Example 1.3: Rewrie he funcion y = f(x) = x 3 from (,) o (3,7) as a vecorvalued funcion. Soluion: Any funcion of he form y = f(x) can be rewrien as a vecorvalued funcion by leing x() = and y() = f(). Thus, he funcion y = f(x) = x 3 from (,) o (3,7) can be rewrien as r() =, 3 for 3. Noe ha r() =, and ha r(3) = 3,7. These are vecors whose heads lie a he poins (,) and (3,7) respecively. Example 1.4: Find he domain of r() =,, 1. 3 Soluion: The domain is he larges subse of he real numbers for which all hree componen funcions are defined simulaneously. Noe ha x() = and y() = are defined for all real numbers, bu ha z() = 1 is no defined when = 3. Thus, he domain of r is given by 3 { (, 3) (3, )}. 78
4 Example 1.: Find he domain of r() =, 4 3, e. Soluion: The firs componen x() = requires ha, and he second componen y() = 4 3 requires ha 4 3, or 4. There are no resricions on implied by z() = 3 e. Thus, he domain of r is given by { (, ) (, 4 3 ]}. Example 1.6: Find a vecor valued funcion ha describes he line segmen in R 3 from (1,,) o (3,1, 4). Soluion: Find he direcion vecor: v = 3 1,1 ( ), 4 =,3, 9. Using (1,,) as he iniial poin, we have 1,, +,3, 9 as he line segmen using vecor noaion. As a vecorvalued funcion, we have r() = 1 +, + 3, 9 for 1. Noe ha r() = 1,,, a vecor whose head lies a he poin (1,,), and ha r(1) = 3,1, 4, a vecor whose head lies a he poin (3,1, 4). Example 1.7: Describe r() = cos, sin, for. Soluion: This is a curve in R 3. Look a wo of he componens a a ime: The componens x() = cos and y() = sin race a circle of radius repeaedly since increases wihou bound. The componens x() = cos and z() = race a cosine wave upward, e.g. assuming ha x is he horizonal axis and z he verical axis. The componens y() = sin and z() = race a sine wave upward. The curve is a helix, which looks like a coiled spring. This helix has a radius of cenered around he posiive zaxis, wrapping around he zaxis (bu never ouching i) as increases in value. 79
5 Example 1.8: In R 3, he circular cylinder x + y = is inerseced by he plane y + z = 4. Find a vecorvalued funcion r() = x(), y(), z() ha describes he curve formed by he inersecion of hese wo surfaces. Soluion: There are many possible vecorvalued funcions ha describe his curve. One possible way is o noe ha we can wrie x() = cos and y() = sin for π. Then, since y + z = 4, we have z = 4 y, so ha z() = 4 sin. The curve of inersecion is given by r() = cos, sin, 4 sin, for π. The number of parameer variables of a vecorvalued funcion describe he ype of graph ha will resul. For example, a vecorvalued funcion of one parameer variable will resul in a curve, as demonsraed in he previous examples. A vecorvalued funcion of wo variables resuls in a surface, as he nex wo examples show. Example 1.9: A circular cylinder of radius is cenered a he origin such ha he xaxis is he axis of symmery of he cylinder. Describe his surface paramerically, using u and v as he parameer variables. Soluion: Since he xaxis is he axis of symmery, we infer ha he circular cross secions lie on planes parallel o he yzplane. For example, a circle of radius on he yzplane (x = ) is described by y + z = 4. Using parameer variable u, we can describe he circle by leing y = cos u and z = sin u, where he represens he circle s radius. Noe ha he circular crosssecions depend only on variable u. Thus, we can le x = v, represening he exension of he circle ino he posiive and negaive x direcion, wih no resricions on v. The cylinder is described paramerically as r(u, v) = v, cos u, sin u, u π, < v <. Example 1.1: Describe he cone z = x + y paramerically using variables u and v. Soluion: Observe ha cross secions of his surface wih a plane z = k resuls in a circle of radius k. Thus, if we le z = u, we can hen define x = u cos v and y = u sin v, which resul in circles of radius u. Thus, we have r(u, v) = u cos v, u sin v, u, where v π and u. 8
6 16. Vecor Valued Funcions: Limis & Coninuiy The same noions of limis and coninuiy hold rue for vecorvalued funcions. For example, he limi of r() = x(), y(), z() as a is given by assuming ha all hree limis exis. lim r() = lim x(), lim y(), lim z(), a a a a Similarly, a vecorvalued funcion r() = x(), y(), z() is coninuous a = a if The limi as a exiss, The vecor r(a) exiss (ha is, a is in he domain of r), and lim r() = r(a). a Example 16.1: Given r() =, e, 1 +3, find lim r(). Is r coninuous a =? Soluion: The limi is lim r() = lim, lim e, lim ( 1 ) = 4, +3 e, 1. Noe ha r() = 4, e, 1. Since all hree condiions of coninuiy are me, he curve raced ou by r() =, e, 1 +3 is coninuous a =. In his example, he limi of r as 3 does no exis since he limi fails o exis for he expression 1. This curve is no coninuous when = 3. I is coninuous everywhere else. +3 Example 16.: Given r() = + 1, 9 3,, find lim 3 r(). Is r coninuous a = 3? Soluion: Noe ha he domain of r excludes he value = 3. However, he limi does exis as 3, since lim r() = lim ( + 1), lim ( 9 ), lim = 7,6,9. The middle expression simplifies as 9 = (+3)( 3) = + 3, hen he limi is aken. However, 3 3 he value = 3 is sill excluded from he domain, so r is no coninuous a = 3. There is a deleed poin in he curve when = 3. 81
7 17. Vecor Valued Funcions: Differeniaion Given a vecorvalued funcion r() = x(), y(), z(), he derivaive of r wih respec o is given by r () = d d r() = d x(), y(), z() d = d d x(), d d y(), d d z() = x (), y (), z (), assuming ha he derivaives exis. Noe ha r () = x (), y (), z () is iself a vecorvalued funcion. Visually, he vecors given by r () can be shifed in such a way so ha hey are angen o he curve raced ou by r() = x(), y(), z(). In a physical seing, if r() = x(), y(), z() represens he displacemen of an objec, hen v() = r () = x (), y (), z () represens he objec s velociy and he magniude, r (), is he objec s speed. Acceleraion is a() = v () = r () = x (), y (), z (). Example 17.1: An objec moves hrough R 3 along a pah defined by r() = 3, +, where all dimensions are in meers. Find he objec s velociy and is speed when = 4 seconds. Soluion: The derivaive of r() = 3, +, is r () = 3, 4 + 1,. Thus, when = 4 seconds, he objec has a velociy of r (4) = 3(4), 4(4) + 1, = 48,17,. The objec s speed a = 4 seconds is r (4) = meers per second. Example 17.: An objec moves hrough R along a pah defined by r() =, , where he firs componen is he horizonal displacemen in meers, and he second componen is verical displacemen in meers, and where is in seconds. Find he maximum heigh ha his objec achieves. Soluion: Noe ha he objec races a downwardopening parabolic arc in R. The objec will achieve is maximum heigh when he verical componen of velociy of he objec is emporarily. Thus, we differeniae: v() = r () = 1,
8 We hen se he verical componen of velociy o, and solve: = gives = seconds. 9.8 This is he ime a which he objec achieves is maximum heigh. When we subsiue =.449 ino r, we have r(.449) =.449, 4.9(.449) + 4(.449) =.449, The objec achieves a maximum heigh of abou meers above he ground afer.449 seconds in fligh. The objec has moved.449 meers horizonally in his same period of ime. Example 17.3: An objec moves hrough R along a pah defined by r() = 3, +, where he componens are in meers and is in seconds. Wha is he minimum speed of he objec? Soluion: The derivaive is r () = 3, +, so ha he speed can be now saed as a funcion in variable : s() = r () = (3 ) + ( + ) = We now minimize s(): d d s() = d d = This expression is when he numeraor is. Using a calculaor, we find ha = when =.48 seconds. This can be verified o be a minimum by using eiher he firs or second derivaive es. Thus, he objec s minimum speed occurs when =.48 seconds and is s(.48) = 9(.48) 4 + 4(.48) + 8(.48) meers per second. Example 17.4: An objec moves hrough R along a pah defined by r() = + 1, 4. Find he equaion of he angen line in vecor form when =. Soluion. The derivaive is r () = 4, 4 3. Thus, when =, he objec is moving (insananeously) in he direcion of r () = 4(), 4() 3 =,. This is he objec s direcion vecor. Furhermore, a =, he objec s locaion is r() = () + 1, () 4 = 1, 6. Thus, he objec s angen line in vecor form when = is 1, 6 +,, or equivalenly, 1 +,
9 Example 17.: An objec moves hrough R 3 along a pah defined by r() = + 3, +,. Find he equaion of he angen line o his pah when he objec is a (7,,). Soluion. As in he previous example, we need boh a direcion vecor and a posiion vecor. The locaion (7,,) corresponds o a posiion vecor 7,,, and seing his equal o r() = + 3, +,, we can deduce ha = 4. The derivaive is r () = 1, + 1,, so he direcion vecor is r (4) = 1, (4) + 1, = 1,9,. Thus, he objec s angen line in vecor form a his insan is 7,, + 1, 9,, or equivalenly, 7 +, + 9, +. Example 17.6: An objec revolves around he origin in a circular orbi. The circle is of radius meers and he objec complees a revoluion every 1 seconds. Assume he objec moves counerclockwise and ha is sared on he posiive xaxis. Find his objec s posiion (displacemen), velociy, speed and acceleraion a ime. Soluion: Le s assume ha 1 seconds represens one revoluion of he objec. Then, he objec s displacemen is given by r() = cos ( π ), sin (π) = cos (π), sin (π). The leading coefficien represens he radius, and noe ha when = 1, he argumens wihin he sine and cosine operaors are boh π (1) = π, he usual period of he sine and cosine funcions. 1 1 The velociy is v() = r () = sin ( π ) (π ), cos (π ) (π ) = π sin (π), π cos (π ), where he chain rule was used followed by simplificaion. Noe ha r() v() =. This is always rue for objecs moving in a circular pah: he (angenial) velociy vecor is orhogonal o he displacemen vecor. The objec s speed is v() = r () = ( π sin ( π )) + (π cos ( π )) = π meers per second. This makes sense: he circumference of he objec s pah is π() = 1π meers. If i akes he objec 1 seconds o complee one revoluion a π meers per second, hen i will have ravelled a disance of 1π meers in ha revoluion. The acceleraion is a() = v () = r () = π cos (π ), π sin (π). Noe ha he acceleraion vecor is always opposie he displacemen vecor for an objec in circular moion. 84
10 18. Vecor Valued Funcions: Inegraion Given a vecorvalued funcion r() = x(), y(), z(), he indefinie inegral of r wih respec o is given by r() d = x() d, y() d, z() d + a, b, c, where a, b, c is a vecor composed of he consans of inegraion of he componens of r. Example 18.1: Find r() d, where r() = 3, 1, sin(3), where >. Soluion: We have r() d = 3 d, ( 1 ) d, sin(3) d = 3, ln, 1 cos(3) + a, b, c. 3 Example 18.: Find r() = r () d, where r () = e,, sin, and r() =,,. Soluion: Noe ha r() = r () d + k, where k = a, b, c is a consan vecor. We have r() = r () d = e d, d, sin() d + k = 1 e, 3 3, cos + a, b, c. Since r() =,,, we have,, = 1 e(), 3 ()3, cos() + a, b, c,, = 1,, 1 + a, b, c. This forces a = 1, b = and c = 1. Thus, r() = 1 e, 3 3, cos + 1,,1, or simplified as r() = 1 (e 1), 3 3, 1 cos. Don confuse r() =,, as being he consan vecor a, b, c. 8
11 Example 18.3: An objec s acceleraion is given by a() =,, where is in seconds and he componens are meers per secondssquared. Find v() and r() such ha v(1) =, and r(1) = 1,3. Soluion: Inegraing acceleraion, we obain velociy: v() = a() d =, d = k 1, 1 + k. To find k = k 1, k, noe ha v(1) =, :, = k 1, 1 (1) + k. This forces k 1 = and k = 9, so ha v() =, Nex, we have r() = v() d =, d = + m 1, m. To find m = m 1, m, we noe ha r(1) = 1,3 : 1,3 = (1) + m 1, 1 6 (1)3 + 9 (1) + m. This forces m 1 = 3 and m = 3. Therefore, r() = 3, Example 18.4: Find r() d, where r() =, e,. +1 Soluion: Inegrae. Noe ha udu subsiuion is used for he laer wo componens. r() d = d = [ ], e d, [ 1 e ] = 8 3, e4 1, 1 ln., ( + 1 ) d, [ 1 ln( + 1)] 86
12 19. Arc Lengh Suppose he vecorvalued funcion r() = x(), y(), z() is defined over he closed inerval a b and differeniable over he open inerval a < < b. Visually, his means ha r is a smooh curve, wih no disconinuiies or corners. The arc lengh s of he curve r over he inerval a b is given by he definie inegral b s = (x ()) + (y ()) + (z ()) d. a Noe ha he inegrand (x ()) + (y ()) + (z ()) is he same as r (). Thus, we can wrie he inegral as b s = r () d. a Example 19.1: Find he lengh of he curve raced by r() = cos, sin for π. Soluion: Find he derivaive: r () = sin, cos. Then, using he arc lengh formula, we have π s = ( sin ) + ( cos ) d π = 4 sin + 4 cos d π = 4(sin + cos ) d π = d = π. The arc lengh is π unis. This can be verified using geomery: r races a semicircle of radius. The circumference of a circle of radius is π() = 4π, and half of his figure is π. 87
13 Example 19.: Find he arc lengh of he curve raced by r() = 4,, ln beween he poins (8,8, ln ) and (,, ln ). Soluion: The derivaive is r () = 4,4,. Furhermore, he bounds of can be inferred from he poins. The poin (8,8, ln ) suggess ha = and he poin (,, ln ) suggess ha =. We have s = 4 + (4) + ( ) d = d = d = (4 + ) d = ( 4 + ) d = (4 + ) d = [ + ln ] = ( + ln ) (8 + ln ) = 4 + ln ( ) unis. Example 19.3: Find he arc lengh of he curve raced by r() =, 3, 4 3 for 1 3. Soluion. The derivaive is r () =, 3,1. Thus, he arc lengh is given by 3 s = () (1 ) d 1 3 = d. 1 Using a calculaor or any numerical mehod of inegraing, we find ha he arc lengh is d unis. 88
14 Example 19.4: Find he lengh of he helix raced by r() = cos, sin, for π. Soluion: The derivaive is r () = sin, cos, 1. We have π s = ( sin ) + ( cos ) + 1 d π = 4 sin + 4 cos + 1 d π = 4(sin + cos ) + 1 d π = d = π unis. Arc Lengh as a Funcion b Consider he arc lengh formula, s = r () d, and allow he upper bound o be a variable a raher han a fixed value. If we allow he upper bound o be, and use a dummy variable wihin he inegral, we have arc lengh s as a funcion of : s() = r (u) du. a Differeniaing boh sides wih respec o, we have d d s() = d d r (u) du. Using he Fundamenal Theorem of Calculus, we have Thus, we have d d r (u) du = r (). a ds d = r (), or equivalenly, ds = r () d. This formula is exremely useful! Do no forge i! a 89
15 . Uni Tangen and Uni Normal Vecors Consider an objec ha moves along a differeniable (smooh, no disconinuiies) curve raced by r() = x(), y(), z(). A each poin on he curve, he angen vecor is given by r () = x (), y (), z (). The magniude of he angen vecor, r (), can be inerpreed as he objec s speed. For mos curves, no surprisingly, he speed of an objec can vary. In a rough sense, he speed of an objec dicaes he segmenaion of he curve. Example.1: Skech he curve raced by r() =, for 4. Soluion: The curve is shown below. I is a parabola y = x from (,) o (4,16). The values for ineger values of are shown on he graph. The segmens of he curve beween consecuive ineger values of vary in lengh. If is a uni of ime, hen he objec raverses each segmen in he same amoun of ime. Thus, he objec mus move faser in order o raverse longer segmens. The segmenaion of he curve in erms of a uni ime inerval is no consisen. The able below shows he objec s posiion, velociy and speed for ineger values of : r() =, r () = 1, r () = 1 + 4, 1, 1 1 1,1 1,,4 1, ,9 1, ,16 1,8 6 9
16 To conrol he speed of he objec, we can force all angen vecors o have a lengh of 1 uni. This is called he uni angen vecor, and is given by This means ha T() = 1. Example.: Find T(), where r() =,. T() = r () r (). Soluion: From he previous example, we have r () = 1, and r () = Thus, T() = r () r () = 1, = , You should verify ha T() = 1. If he objec moves along his curve a a consan speed of 1 uni of disance per uni of ime, hen his will force he segmenaion of he curve ino equalsized segmens, so ha i can raverse he same lengh each ime, per uni of ime. This is ofen called he ds segmenaion. Example.3: Find T(), where r() = 3 cos, 3 sin,. Soluion: We have T() = r () 3 sin, 3 cos, 1 3 sin r = = () cos, 1, 1 1. Noe ha in his case, he speed of he objec is always 1 unis of disance per uni of ime. The uni normal vecor is given by N() = T () T (). The vecor N has a lengh of 1 uni. I is orhogonal o T (ha is, N T = ). For an objec moving along a differeniable curve, T will poin in he objec s (angenial) direcion of ravel, and N will poin orhogonal o T, represening one componen of acceleraion. I generally poins inward o concave side of he curve. 91
17 Example.4: Find N(), where r() =,. Soluion: From Example., we have 1 T() = 1 + 4, We now find T (): T 4 () = (1 + 4 ) 3, (1 + 4 ) 3. Now, we need T () : This simplifies afer many seps o Thus, he uni normal N is given by T 4 () = ( (1 + 4 ) 3 ) + ( (1 + 4 ) 3 T () = ) N() = T () T () = 1 4 ( ) (1 + 4 ) 3, (1 + 4 ) 3 = 1 + 4, Noe he similariies in T and N, and noe also ha N T =. Example.: Find N(), where r() = 3 cos, 3 sin,. Soluion: From Example.3, we have We find T (): 3 sin T() = 1 3 cos T () = 1, 3 cos 1, 1 1., 3 sin 1,.. 9
18 Noe ha T 3 cos () = ( 1 ) 3 sin + ( 1 ) = 3 1. Thus, 3 cos 3 sin N() = T (), T () = 1 1, 3 1 Observe ha N() = 1 and ha N T =. = cos, sin,. See an error? Have a suggesion? Please see 93
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