Lecture XVII. Vector functions, vector and scalar fields Definition 1 A vector-valued function is a map associating vectors to real numbers, that is

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Lecture XVII Abstrct We introduce the concepts of vector functions, sclr nd vector fields nd stress their relevnce in pplied sciences. We study curves in three-dimensionl Eucliden spce nd introduce the concept of rc length, curvture nd torsion for given curve in R 3. Vector functions, vector nd sclr fields Definition 1 A vector-vlued function is mp ssociting vectors to rel numbers, tht is r : R R 3 such tht r(t) = f(t)e x + g(t)e y + h(t)e z, where e x, e y, nd e z re the unit vectors long the x, y, nd z-xes, respectively. The domin of r is the intersection of the domins of the functions f, g nd h. Notice tht if the prmeter t is interpreted s time, then we cn think to r(t) s function describing the trjectory of prticle of mss m in the three-dimensionl Eucliden spce. Indeed, we could obtin r(t) by solving Newton 1 s eqution m d2 r dt 2 = F, where F is the externl force cting on the prticle. Exmple 1 Find the domin of definition of the vector-vlued function Notice tht r(t) = t 3 e x + ln (3 t)e y + te z. t 3 is polynomil nd its domin of definition is the whole rel line. ln (3 t) will be defined if 3 t > 0, tht is for t < 3. t is defined if t 0. Hence, we conclude tht the domin of definition of the given vector-vlued function is 0 t < 3, tht is the intervl [0, 3). 1 Isc Newton (1643-1727) ws n English physicist, mthemticin, stronomer, nturl philosopher, lchemist nd theologin. 1

In mthemtics vector field is construction in vector clculus which ssocites vector to every point in subset of the Eucliden spce R 3, tht is mp V : R 3 R 3 such tht V(x, y, z) = F (x, y, z)e x +G(x, y, z)e y +H(x, y, z)e z. Vector fields re often used in physics nd engineering to model, for exmple, the speed nd direction of moving fluid throughout spce, or the strength nd direction of some force, such s the mgnetic, electric or grvittionl force, s it chnges from point to point. In mthemtics, physics nd engineering sclr field ssocites sclr vlue to every point in spce, tht is mp S : R 3 R such tht (x, y, z) S(x, y, z). Sclr fields re required to be coordinte-independent, mening tht ny two observers using the sme units will gree on the vlue of the sclr field t the sme point in spce (or spcetime). Exmples used in physics nd engineering include the temperture distribution throughout spce, the pressure distribution in fluid, the density of fluid or solid object, the electrosttic potentil, etc.. Definition 2 We define the limit of vector-vlued function r(t) s t goes to R s lim r(t) = lim f(t) e x + lim g(t) e y + lim h(t) e z. t t t t If the bove limit exists we sy tht r(t) is continuous t t =. Exmple 2 Find the limit for t 1 of r(t) = t + 3 e x + t 1 t 2 1 e y + tn t t According to the definition bove we hve to compute the corresponding limits for the components of the given vector-vlued function. Thus, we find tht lim t + 3 = 4 = 2, t 1 t 1 lim t 1 t 2 1 = lim t 1 tn t lim t 1 t Hence, we conclude tht t 1 (t 1)(t + 1) = lim t 1 = tn(1) 1.557. lim r(t) = 2e x + e y t 1 2 + tn (1) e z. 2 e z. 1 t + 1 = 1 2,

In mthemtics, prmetric eqution is reltion defined by the use of prmeters. A simple kinemticl exmple is when we use time prmeter to determine the position, velocity, nd other informtion bout body or prticle in motion. In generl, prmetric eqution is expressed by mens of set of equtions. For exmple, the simplest eqution for prbol y = x 2 cn be prmetrized by using free prmeter t, nd setting x = t, y = t 2. Although the preceding exmple is somewht trivil cse, consider for instnce the following prmetriztion of circle of rdius R x = R cos t, y = R sin t, where t is in the rnge 0 to 2π. Converting set of prmetric equtions to single eqution involves eliminting the vrible t from the simultneous equtions x = x(t) nd y = y(t). If one of these equtions cn be solved for t, the expression obtined cn be substituted into the other eqution to obtin n eqution involving x nd y only. In the cse of the exmple with the prmetric equtions of circle we cn squre both x nd y, dd them together nd end up with the well-known formul x 2 + y 2 = R 2. Notice tht once we hve vector-vlued function r(t) = f(t)e x + g(t)e y + h(t)e z describing the trjectory of prticle in R 3 the curve on which this prticle moves cn lso be described in prmetric form by setting x = f(t), y = g(t), z = h(t). Notice tht t ech time t the bove eqution will fix certin point of coordintes (x, y, z) in spce. Exmple 3 Consider the vector-vlued function r(t) = cos t e x + sin t e y + te z. 3

We wnt to understnd the form of the trjectory whose points re described by the bove vector-vlued function s the prmeter t vries. If we denote by C such trjectory, the prmetric equtions for the curve C re x = cos t, y = sin t, z = t. Since x 2 + y 2 = 1 whtever is the vlue of z we conclude tht the points of the curve C must ly on the lterl surfce of cylinder of rdius one. Furthermore, if we consider for instnce different vlues of t, for instnce we strt by tking t = 0, t = π/2, t = 3/2π nd so on we discover tht the trjectory of the prticle is helix. For picture downlod the file helix.gif on the web pge of the course. The plot hs been generted with the commnd spcecurve in Mple nd by tking t in the intervl [ 6π, 6π]. Definition 3 Let r(t) be the vector-vlued function r(t) = f(t)e x + g(t)e y + h(t)e z. We define the derivtive r (t) of r(t) s r(t + t) r(t) r (t) = lim. t 0 t If the bove limit exists, then we hve r (t) = df dt e x + dg dt e y + dh dt e z. Concerning the geometricl interprettion of r (t), if r(t) describes curve C in R 3 the derivtive r (t) is the tngent vector to the curve C t the point P = (x(t), y(t), z(t)). In wht follows we shll denote the unit tngent vector by T(t) = r (t) r (t). From the physicl point of view r (t) cn be interpreted s the velocity of prticle hving trjectory C described by the vector-vlued function r(t). The interprettion of the second derivtive of r(t) is tht of n ccelertion. If nd denote the dot nd cross product, respectively, we hve the following differentition rules d dt (u v) = u v + u v, d dt (u v) = u v + u v, d dt [u(f(t))] = f (t)u (f(t)), (chin rule) 4

where u(t) nd v(t) re two rbitrry vector-vlued functions. Definition 4 Let r(t) = f(t)e x + g(t)e y + h(t)e z. be continuous vector-vlued function. We define the integrl of r(t) s follows b ( b ) ( b ) ( b ) dt r(t) = dt f(t) e x + dt g(t) e y + dt h(t) e z. A curve in the plne cn be pproximted by connecting finite number of points on the curve using line segments to crete polygonl pth. Since it is strightforwrd to clculte the length of ech liner segment (using the Pythgoren theorem in Eucliden spce, for exmple), the totl length of the pproximtion cn be found by summing the lengths of ech liner segment. If the curve is not lredy polygonl pth, better pproximtions to the curve cn be obtined by following the shpe of the curve incresingly more closely. The pproch is to use n incresingly lrger number of segments of smller lengths. The lengths of the successive pproximtions do not decrese nd will eventully keep incresingpossibly indefinitely, but for smooth curves this will tend to limit s the lengths of the segments get rbitrrily smll. For some curves there is smllest number L tht is n upper bound on the length of ny polygonl pproximtion. If such number exists, then the curve is sid to be rectifible nd the curve is defined to hve rc length L. Definition 5 Let C be curve in R 3 described by the vector-vlued function r(t) = x(t)e x + y(t)e y + z(t)e z where t [, b]. If C is described only once s t increses from to b, then its length or rc length is given by the formul b b (dx ) 2 ( ) 2 ( ) 2 dy dz L = dt r (t) = dt + +. dt dt dt Exmple 4 Find the length of the helix r(t) = cos t e x + sin t e y + t e z from the point (1, 0, 0) to (1, 0, 2π). First of ll, we hve to find the vlues of t corresponding to the two given points. For the first point we obtin 1 e x + 0 e y + 0 e z = cos t e x + sin t e y + t e z. 5

The bove vectoril eqution gives rise to the following three equtions cos t = 1, sin t = 0, t = 0. which re clerly stisfied for t = 0. In n nlogous wy we find tht r(t) will identify the point (1, 0, 2π) whenever t = 2π. Furthermore, Hence, the length will be given by r (t) = sin t e x + cos t e y + e z, r (t) = sin 2 t + cos 2 t + 1 = 2. L = 2π 0 dt r (t) = 2π 0 dt 2 = 2π 2. It is often convenient to prmetrize curve with respect to rc length becuse rc length rises from the shpe of curve nd does not depend on prticulr coordinte system. Moreover, with this reprmeteriztion we cn tell where we re on the curve fter we hve trvelled distnce s long the curve. Definition 6 Let C be curve in R 3 described by the vector-vlued function r(t) = x(t)e x + y(t)e y + z(t)e z where t [, b]. Further suppose tht C is described only once s t increses from to b. We define the rc length function s of the curve C s t t (dx ) 2 ( ) 2 ( ) 2 dy dz s(t) = dτ r (τ) = dτ + +. (1) dτ dτ dτ Notice tht from the Fundmentl Theorem of Clculus we lso hve ds dt = r (t). (2) Exmple 5 Reprmeterize the helix r(t) = cos t e x + sin t e y + t e z with respect to rc length from (1, 0, 0) in the direction of incresing t. We lredy know tht the initil point is ssocited to t = 0. Computing the integrl in (1) we obtin s(t) = We conclude tht t 0 r(s) = cos dτ r (τ) = 2 t 0 dτ = 2t = t = ( ) ( ) s s 2 e x + sin 2 e y + s e z. 2 s 2. 6

The curvture of curve C cn be thought s mesure of how quickly the curve chnges direction t given point. Reclling tht the direction t given point on curve is given by the unit tngent vector T t tht point it results tht the curvture will be connected to the rte of chnge of T long C. This intuitive ide is mde rigorous in the following definition. Definition 7 Let C be curve in R 3 described by vector-vlued function r(t). The curvture of C is κ = dt ds, T = r (t) r(t). Even though the bove formul reflects our heuristic ide of curvture of curve it is not well suited for prcticl purposes. In fct, it results to be more convenient to derive formul for the curvture involving only derivtives with respect to t. This cn be chieved by observing tht dt ds = dt dt dt ds = dt dt ds dt = T r, where we used (2) in the lst step. Hence, we found n equivlent formul for the curvture given by κ = T r. (3) Exmple 6 Show tht the curvture of circle of rdius R is κ = R 1. First of ll, points lying on circle of rdius R cn be described by vector-vlued function r(t) = R cos t e x + R sin t e y. In order to pply formul (3) we need to compute the following quntities r (t) = R sin t e x + R cos t e y, r (t) = R 2 sin 2 t + R 2 cos 2 t = R, T = r (t) r (t) = sin t e x + cos t e y, T = 1. Hence, we conclude tht κ = T r = 1 R. 7

Finlly, very useful formul for the computtion of curvture involving only r(t) nd its first nd second derivtives is κ = r (t) r (t). (4) r (t) 3 Exmple 7 Find the curvture of the curve described by r(t) = t e x + t 2 e y + t 3 e z. In order to pply (4) we hve to compute r (t) = e x + 2t e y + 3t 2 e z = r (t) 3 = (1 + 4t 2 + 9t 4 ) 3/2, r (t) = 2 e y + 6t e z. On the other side r (t) r (t) = det e x e y e z 1 2t 3t 2 0 2 6t = 6t 2 e x 6t e y + 2 e z nd Thus, we conclude tht r (t) r (t) = 2 9t 4 + 9t 2 + 1. κ(t) = 2 9t 4 + 9t 2 + 1 (1 + 4t 2 + 9t 4 ) 3/2 = 2 9t 4 + 9t 2 + 1 (1 + 4t 2 + 9t 4 ) 3. Exmple 8 We wnt to rewrite the formul (4) for the specil cse of plne curve y = y(x). Clerly, points on this curve hve coordintes (x, y(x)) nd re identified by position vector Moreover, r(x) = x e x + y(x) e y. r (x) = e x + y (x) e y = r (x) 3 = [1 + (y ) 2 ] 3/2, r (x) = y e y. On the other side r (x) r (x) = det e x e y e z 1 y 0 0 y 0 8 = y (x) e z

nd Thus, we conclude tht r (x) r (x) = y. κ(x) = y [1 + (y ) 2 ] 3/2. The torsion of curve mesures how shrply it is twisting nd if r(t) is vector-vlued function describing curve in spce the torsion τ of the curve cn be computed with the formul τ = [r (t) r (t)] r (t). r (t) r (t) 2 It is not difficult to check tht the torsion of helix is constnt. Prctice problems 1. Find the domin of definition of the vector functions () r(t) = t 2 e x + t 1 e y + 5 t e z ; (b) r(t) = ln t e x + t t 1 e y + e t e z. 2. Find () the limit for t 0 + of (b) the limit for t 0 of r(t) = cos t e x + sin t e y + t ln t e z ; r(t) = et 1 t e x + 1 + t 1 t e y + 3 1 + t e z. 3. The eqution of cone in R 3 is given by z 2 = x 2 + y 2. Show tht the curve with prmetric equtions x = t cos t, y = t sin t nd z = t lies on tht cone nd use this fct to help sketch the curve. 4. Show tht the curve with prmetric equtions x = t 2, y = 1 3t, nd z = 1 + t 3 psses through the points (1, 4, 0) nd (9, 8, 28) but not through the point 4, 7, 6). 9

5. Find vector vlued-function representing the curve of intersection of the circulr cylinder {(x, y, z) R 3 x 2 +y 2 = 1, z R} nd the plne y + z = 2. 6. Find the derivtive of the following vector-vlued functions () r(t) = e x e y + e 4t e z, (b) r(t) = e t cos t e x + e t sin t e y + ln t e z, (c) r(t) = + t b + t 2 c, (d) r(t) = t (b + t c), where, b, nd c re rbitrry constnt vectors in R 3 7. Find the unit tngent vector T t the point with the given vlue of the prmeter t () r(t) = t e x + (t t 2 ) e y + rctn t e z, t = 1, (b) r(t) = e 2t cos t e x + e 2t sin t e y + e 2t e z, t = π 2. 8. Evlute the integrls 9. If () (b) find 4 1 π/4 0 ( dt t ex + te t e y + e z ), t 2 dt [cos (2t) e x + sin (2t) e y + t sin t e z ]. u(t) = e x 2t 2 e y + 3t 3 e z, v(t) = t e x + cos t e y + sin t e z, 10. If r(t) 0, show tht d dt [u(t) v(t)] nd d [u(t) v(t)]. dt d dt r(t) = r(t) r (t). r(t) 11. Reprmetrize the curve with respect to rc length mesured from the point where t = 0 in the direction of incresing t () r(t) = e t sin t e x + e t cos t e y, 10

(b) r(t) = (1 + 2t) e x + (3 + t) e y 5t e z, (c) r(t) = 3 sin t e x + 4t e y + 3 cos t e z. 12. Reprmetrize the curve ( ) 2 r(t) = t 2 + 1 1 e x + 2t t 2 + 1 e y with respect to rc length mesured from the point (1, 0) in the direction of incresing t. Express the reprmetriztion in its simplest form. Wht cn you conclude bout the curve? 13. Find the curvture of the curves described by the vector-vlued functions () r(t) = t 2 e x + (sin t t cos t) e y + (cos t + t sin t) e z, t > 0, (b) r(t) = t 2 e x + 2t e y + ln t e z, t > 0, 14. Find the curvture of r(t) = 2t e x +e t e y +e t e z t the point (0, 1, 1). 15. Grph the curve with prmetric equtions x = t, y = 4t 3/2, z = t 2 nd find the curvture t the point (1, 4, 1). 16. Find the curvture of y = x 3, y = x, y = sin x. 17. At wht point does the curve hve mximum curvture? Wht hppens to the curvture s x? y = ln x, y = e x. 18. Find n eqution of prbol tht hs curvture 4 t the origin. 19. Show tht the circulr helix r(t) = cos t e x + sin t e y + bt e z where nd b re positive constnts, hs constnt curvture nd constnt torsion. 20. Find the curvture nd torsion of the curve x = sinh t, y = cosh t, z = t t the point (0, 1, 0). 11

21. The DNA molecule hs the shpe of double helix. The rdius of ech helix is bout 10 ngstroms (1 ngstrom = 10 8 cm). Ech helix rises bout 34 ngstroms during ech complete turn, nd there re bout 2.9 10 8 complete turns. Estimte the length of ech helix. 22. We consider the problem of designing rilrod trck to mke smooth trnsition between sections of stright trck. Existing trck long the negtive x-xis is to be joined smoothly to trck long the line y = 1 for x 1. Find polynomil P = P (x) of degree 5 such tht the function defined by 0 if x 0, F (x) = P (x) if 0 < x < 1, 1 if x 1, is continuous, hs continuous slope nd continuous curvture. 23. Wht force is required so tht prticle of mss m hs the position function r(t) = t 3 e x + t 2 e y + t 3 e z? 24. A force of mgnitude 20 N cts directly upwrd from the xy-plne on n object of mss 4 Kg. The object strts t the origin with initil velocity v(0) = e x e y. Find its position function nd its speed t time t. 25. A projectile is fired with n initil speed of 500 m/s nd ngle of elevtion 30 o. Find the rnge of the projectile, the mximum height reched nd the speed t impct. 26. The position function of spceship is r(t) = (3 + t) e x + (2 + ln t) e y + ( 7 4 ) t 2 + 1 nd the coordintes of spce sttion re (6, 4, 9). The cptin wnts the spceship to cost into the spce sttion. When should the engines be turned off? 27. A rocket burning its onbord fuel while moving through spce hs velocity v(t) nd mss m(t) t time t. If the exhust gses escpe with velocity v e reltive to the rocket, it cn be deduced from Newton s Second Lw of Motion tht m dv dt = dm dt v e. 12 e z

() Show tht v(t) = v(0) ( ln m(0) ) v e. m(t) (b) For the rocket to ccelerte in stright line from rest to twice the speed of its own exhust gses, wht frction of its initil mss would the rocket hve to burn s fuel? 13