Vector Functions. Exercises Chapter 13 Vector Functions

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1 33 Chpter 13 Vector Functions 13 Vector Functions shown for t between n 2π Both strt n en t the sme point, but the first helix tkes two full turns to get there, becuse its z coorinte grows more slowly ½ º½ ËÔ ÙÖÚ We hve lrey seen tht convenient wy to escribe line in three imensions is to provie vector tht points to every point on the line s prmeter t vries, like 1,2,3 +t 1, 2,2 = 1+t,2 2t,3+2t Except tht this gives prticulrly simple geometric object, there is nothing specil bout the iniviul functions of t tht mke up the coorintes of this vector ny vector with prmeter, like f(t), g(t), h(t), will escribe some curve in three imensions s t vries through ll possible vlues EXAMPLE 1311 Describethecurves cost,sint,, cost,sint,t,n cost,sint,2t As t vries, the first two coorintes in ll three functions trce out the points on the unit circle, strting with (1, ) when t = n proceeing counter-clockwise roun the circlestincreses Inthefirst cse, thez coorinteislwys, so thisescribes precisely the unit circle in the x-y plne In the secon cse, the x n y coorintes still escribe circle, but now the z coorinte vries, so tht the height of the curve mtches the vlue of t When t = π, for exmple, the resulting vector is 1,,π A bit of thought shoul convince you tht the result is helix In the thir vector, the z coorinte vries twice s fst s the prmeter t, so we get stretche out helix Both re shown in figure 1311 On the left is the first helix, shown for t between n 4π; on the right is the secon helix, Figure 1311 Two helixes (AP) A vector expression of the form f(t), g(t), h(t) is clle vector function; it is function from the rel numbers R to the set of ll three-imensionl vectors We cn lterntely think of it s three seprte functions, x = f(t), y = g(t), n z = h(t), tht escribe pointsinspce Inthiscsewe usullyrefer tothe set ofequtionssprmetric equtions for the curve, just s for line While the prmeter t in vector function might represent ny one of number of physicl quntities, or be simply pure number, it is often convenient n useful to think of t s representing time The vector function then tells you where in spce prticulr object is t ny time Vector functions cn be ifficult to unerstn, tht is, ifficult to picture When vilble, computer softwre cn be very helpful When working by hn, one useful pproch is to consier the projections of the curve onto the three stnr coorinte plnes We hve lrey one this in prt: in exmple 1311 we note tht ll three curves project to circle in the x-y plne, since cost,sint is two imensionl vector function for the unit circle EXAMPLE 1312 Grph the projections of cost,sint,2t onto the x-z plne n the y-z plne The two imensionl vector function for the projection onto the x-z plne is cost,2t, or in prmetric form, x = cost, z = 2t By eliminting t we get the eqution x = cos(z/2), the fmilir curve shown on the left in figure 1312 For the projection onto the y-z plne, we strt with the vector function sint,2t, which is the sme s y = sint, z = 2t Eliminting t gives y = sin(z/2), s shown on the right in figure π 2π z 1 1 x 132 Clculus with vector functions 331 4π 2π z 1 1 Figure 1312 The projections of cost, sint, 2t onto the x-z n y-z plnes Exercises Describe the curve r = sint,cost,cos8t 2 Describe the curve r = tcost,tsint,t 3 Describe the curve r = t,t 2,cost 4 Describe the curve r = cos(2t) 1 t 2,sin(2t) 1 t 2,t 5 Fin vector function for the curve of intersection of x 2 +y 2 = 9 n y +z = 2 6 A bug is crwling outwr long the spoke of wheel tht lies long rius of the wheel The bug is crwling t 1 unit per secon n the wheel is rotting t 1 rin per secon Suppose the wheel lies in the y-z plne with center t the origin, n t time t = the spoke lies long the positive y xis n the bug is t the origin Fin vector function r(t) for the position of the bug t time t 7 Wht is the ifference between the prmetric curves f(t) = t,t,t 2, g(t) = t 2,t 2,t 4, n h(t) = sin(t),sin(t),sin 2 (t) s t runs over ll rel numbers? 8 Plot ech of the curves below in 2 imensions, projecte onto ech of the three stnr plnes (the x-y, x-z, n y-z plnes) f(t) = t,t 3,t 2, t rnges over ll rel numbers b f(t) = t 2,t 1,t 2 +5 for t 3 9 Given points A = (1,2,3) n B = (b1,b2,b3), give prmetric equtions for the line segment connecting A n B Be sure to give pproprite t vlues 1 With prmetric plot n set of t vlues, we cn ssocite irection For exmple, the curve cost,sint is the unit circle trce counterclockwise How cn we men set of given prmetric equtions n t vlues to get the sme curve, only trce bckwrs? ½ º¾ ÐÙÐÙ Û Ø Ú ØÓÖ ÙÒØ ÓÒ A vector function r(t) = f(t),g(t),h(t) is function of one vrible tht is, there is only one input vlue Wht mkes vector functions more complicte thn the functions y 332 Chpter 13 Vector Functions y = f(x) tht we stuie in the first prt of this book is of course tht the output vlues re now three-imensionl vectors inste of simply numbers It is nturl to woner if there is corresponing notion of erivtive for vector functions In the simpler cse of function y = s(t), in which t represents time n s(t) is position on line, we hve seen tht the erivtive s (t) represents velocity; we might hope tht in similr wy the erivtive of vector function woul tell us something bout the velocity of n object moving in three imensions One wy to pproch the question of the erivtive for vector functions is to write own n expression tht is nlogous to the erivtive we lrey unerstn, n see if we cn mke sense of it This gives us r r(t+) r(t) (t) = lim f(t+) f(t),g(t+) g(t),h(t+) h(t) = lim = lim f(t+) f(t) = f (t),g (t),h (t),, g(t+) g(t), h(t+) h(t) if we sy tht wht we men by the limit of vector is the vector of the iniviul coorinte limits So strting with fmilir expression for wht ppers to be erivtive, we fin tht we cn mke goo computtionl sense out of it but wht oes it ctully men? We know how to interpret r(t+) n r(t) they re vectors tht point to loctions in spce; if t is time, we cn think of these points s positions of moving object t times tht re prt We lso know wht r = r(t +) r(t) mens it is vector tht points from the he of r(t) to the he of r(t + ), ssuming both hve their tils t the origin So when is smll, r is tiny vector pointing from one point on the pth of the object to nerby point As gets close to, this vector points in irection tht is closer n closer to the irection in which the object is moving; geometriclly, it pproches vector tngent to the pth of the object t prticulr point r r(t) r(t+) Figure 1321 Approximting the erivtive

2 132 Clculus with vector functions 333 Unfortuntely, the vector r pproches in length; the vector,, is not very informtive By iviing by, when it is smll, we effectively keep mgnifying the length of r so tht in the limit it oesn t ispper Thus the limiting vector f (t),g (t),h (t) will (usully) be goo, non-zero vector tht is tngent to the curve Wht bout the length of this vector? It s nice tht we ve kept it wy from zero, but wht oes it mesure, if nything? Consier the length of one of the vectors tht pproches the tngent vector: r(t+) r(t) = r(t+) r(t) The numertor is the length of the vector tht points from one position of the object to nerby position; this length is pproximtely the istnce trvele by the object between times t n t + Diviing this istnce by the length of time it tkes to trvel tht istnce gives the verge spee As pproches zero, this verge spee pproches the ctul, instntneous spee of the object t time t So by performing n obvious clcultion to get something tht looks like the erivtive of r(t), we get precisely wht we woul wnt from such erivtive: the vector r (t) points in the irection of trvel of the object n its length tells us the spee of trvel In the cse tht t is time, then, we cll v(t) = r (t) the velocity vector Even if t is not time, r (t) is useful it is vector tngent to the curve EXAMPLE 1321 We hve seen tht r = cost,sint,t is helix We compute r = sint,cost,1, n r = sin 2 t+cos 2 t+1 = 2 So thinking of this s escription of moving object, its spee is lwys 2; see figure Chpter 13 Vector Functions EXAMPLE 1322 The velocity vector for cost,sint,cost is sint,cost, sint As before, the first two coorintes men tht from bove this curve looks like circle The z coorinte is now lso perioic, so tht s the object moves roun the curve its height oscilltes up n own In fct it turns out tht the curve is tilte ellipse, s shown in figure 1323 Figure 1323 The ellipse r = cost, sint, cost (AP) EXAMPLE 1323 The velocity vector for cost,sint,cos2t is sint,cost, 2sin2t The z coorinte is now oscillting twice s fst s in the previous exmple, so the grph is not surprising; see figure 1324 Figure 1324 cost, sint, cos2t (AP) Figure 1322 A tngent vector on the helix (AP) 132 Clculus with vector functions 335 EXAMPLE 1324 Fin the ngle between the curves t,1 t,3+t 2 n 3 t,t 2,t 2 where they meet The ngle between two curves t point is the ngle between their tngent vectors ny tngent vectors will o, so we cn use the erivtives We nee to fin the point of intersection, evlute the two erivtives there, n finlly fin the ngle between them To fin the point of intersection, we nee to solve the equtions t = 3 u 1 t = u 2 3+t 2 = u 2 Solving either of the first two equtions for u n substituting in the thir gives 3+t 2 = (3 t) 2, which mens t = 1 This together with u = 2 stisfies ll three equtions Thus the two curves meet t (1,,4), the first when t = 1 n the secon when t = 2 The erivtives re 1, 1,2t n 1,1,2t ; t the intersection point these re 1, 1,2 n 1,1,4 The cosine of the ngle between them is then cosθ = = 1 3, 336 Chpter 13 Vector Functions Note tht becuse the cross prouct is not commuttive you must remember to o the three cross proucts in formul (e) in the correct orer When the erivtive of function f(t) is zero, we know tht the function hs horizontl tngent line, n my hve locl mximum or minimum point If r (t) =, the geometric interprettion is quite ifferent, though the interprettion in terms of motion is similr Certinly we know tht the object hs spee zero t such point, n it my thus be bruptly chnging irection In three imensions there re mny wys to chnge irection; geometriclly this often mens the curve hs cusp or point, s in the pth of bll tht bounces off the floor or wll EXAMPLE 1326 Suppose tht r(t) = 1+t 3,t 2,1, so r (t) = 3t 2,2t, This is t t =, n there is inee cusp t the point (1,,1), s shown in figure so θ = rccos(1/ 3) 96 The erivtives of vector functions obey some fmilir looking rules, which we will occsionlly nee THEOREM 1325 Suppose r(t) n s(t) re ifferentible functions, f(t) is ifferentible function, n is rel number t r(t) = r (t) b t (r(t)+s(t)) = r (t)+s (t) c t f(t)r(t) = f(t)r (t)+f (t)r(t) t (r(t) s(t)) = r (t) s(t)+r(t) s (t) e t (r(t) s(t)) = r (t) s(t)+r(t) s (t) f t r(f(t)) = r (f(t))f (t) Figure t 3,t 2,1 hs cusp t 1,,1 (AP) 2 Sometimes we will be intereste in the irection of r but not its length In some cses, we cn still work with r, s when we fin the ngle between two curves On other occsions it will be useful to work with unit vector in the sme irection s r ; of course, we cn compute such vector by iviing r by its own length This stnr unit tngent vector is usully enote by T: 15 T = r r In sense, when we compute the ngle between two tngent vectors we hve lrey me use of the unit tngent, since cosθ = r s r s = r r s s 1

3 132 Clculus with vector functions Chpter 13 Vector Functions Now tht we know how to mke sense of r, we immeitely know wht n ntierivtive must be, nmely r(t)t = f(t) t, g(t) t, h(t)t, if r = f(t), g(t), h(t) Wht bout efinite integrls? Suppose tht v(t) gives the velocity of n object t time t Then v(t) is vector tht pproximtes the isplcement of the object over the time : v(t) points in the irection of trvel, n v(t) = v(t) is the spee of the object times, which is pproximtely the istnce trvele Thus, if we sum mny such tiny vectors: n 1 v(t i) i= we get n pproximtion to the isplcement vector over the time intervl [t,t n] If we tke the limit we get the exct vlue of the isplcement vector: n 1 n lim v(t i) = i= t v(t)t = r(t n) r(t ) Denote r(t ) by r Then given the velocity vector we cn compute the vector function r giving the loction of the object: r(t) = r + v(u) u EXAMPLE 1327 An object moves with velocity vector cos t, sint, cost, strting t (1,1,1) t time Fin the function r giving its loction See figure 1326 r(t) = 1,1,1 + cosu,sinu,cosu u = 1,1,1 + sinu, cosu,sinu t t = 1,1,1 + sint, cost,sint, 1, = 1+sint,2 cost,1+sint Exercises 132 Figure 1326 Pth of the object with its initil velocity vector (AP) 1 Fin r n T for r = t 2,1,t 2 Fin r n T for r = cost,sin2t,t 2 3 Fin r n T for r = cos(e t ),sin(e t ),sint 4 Fin vector function for the line tngent to the helix cost,sint,t when t = π/4 5 Fin vector function for the line tngent to cost,sint,cos4t when t = π/3 6 Fin the cosine of the ngle between the curves,t 2,t n cos(πt/2),sin(πt/2),t where they intersect 7 Fin the cosine of the ngle between the curves cost, sin(t)/4, sint n cost, sint, sin(2t) where they intersect 8 Suppose tht r(t) = k, for some constnt k This mens tht r escribes some pth on the sphere of rius k with center t the origin Show tht r is perpeniculr to r t every point Hint: Use Theorem 1325, prt () 9 A bug is crwling long the spoke of wheel tht lies long rius of the wheel The bug is crwling t 1 unit per secon n the wheel is rotting t 1 rin per secon Suppose the wheel lies in the y-z plne with center t the origin, n t time t = the spoke lies long the positive y xis n the bug is t the origin Fin vector function r(t) for the position of the bug t time t, the velocity vector r (t), the unit tngent T(t), n the spee of the bug r (t) 1 An object moves with velocity vector cost,sint,t, strting t,, when t = Fin the function r giving its loction 11 The position function of prticle is given by r(t) = t 2,5t,t 2 16t, t When is the spee of the prticle minimum? 12 A prticle moves so tht its position is given by cost,sint,cos(6t) Fin the mximum n minimum spees of the prticle 13 An object moves with velocity vector t,t 2,cost, strting t,, when t = Fin the function r giving its loction 133 Arc length n curvture Wht is the physicl interprettion of the ot prouct of two vector vlue functions? Wht is the physicl interprettion of the cross prouct of two vector vlue functions? 15 Show, using the rules of cross proucts n ifferentition, tht t (r(t) r (t)) = r(t) r (t) 16 Determine the point t which f(t) = t,t 2,t 3 n g(t) = cos(t),cos(2t),t+1 intersect, n fin the ngle between the curves t tht point (Hint: You ll nee to set this one up like line intersection problem, writing one in s n one in t) If these two functions were the trjectories of two irplnes on the sme scle of time, woul the plnes collie t their point of intersection? Explin 17 Fin the eqution of the plne perpeniculr to the curve r(t) = 2sin(3t),t,2cos(3t) t the point (,π, 2) 18 Fin the eqution of the plne perpeniculr to cost,sint,cos(6t) when t = π/4 19 At wht point on the curve r(t) = t 3,3t,t 4 is the plne perpeniculr to the curve lso prllel to the plne 6x+6y 8z = 1? 2 Fin the eqution of the line tngent to cost,sint,cos(6t) when t = π/4 ½ º Ö Ð Ò Ø Ò ÙÖÚ ØÙÖ Sometimes it is useful to compute the length of curve in spce; for exmple, if the curve represents the pth of moving object, the length of the curve between two points my be the istnce trvele by the object between two times Recll tht if the curve is given by the vector function r then the vector r = r(t+ ) r(t) points from one position on the curve to nother, s epicte in figure 1321 If the points re close together, the length of r is close to the length of the curve between the two points If we up the lengths of mny such tiny vectors, plce he to til long segment of the curve, we get n pproximtion to the length of the curve over tht segment In the limit, s usul, this sum turns into n integrl tht computes precisely the length of the curve First, note tht r = r r (t), when is smll Then the length of the curve between r() n r(b) is n 1 n 1 n 1 r b lim r = lim = lim r (t) = r (t) t n n n i= i= i= (Well, sometimes This works if between n b the segment of curve is trce out exctly once) 34 Chpter 13 Vector Functions EXAMPLE 1331 Let s fin the length of one turn of the helix r = cost,sint,t (see figure 1311) We compute r = sint,cost,1 n r = sin 2 t+cos 2 t+1 = 2, so the length is 2π 2t = 2 2π EXAMPLE 1332 Suppose y = lnx; wht is the length of this curve between x = 1 n x = 3? Although this problem oes not pper to involve vectors or three imensions, we cn interpret it in those terms: let r(t) = t, lnt, This vector function trces out precisely y = lnx in the x-y plne Then r (t) = 1,1/t, n r (t) = 1+1/t 2 n the esire length is t t = 2 2+ln( 2+1) ln3 (This integrl is bit tricky, but requires only methos we hve lerne) Notice tht there is nothing specil bout y = lnx, except tht the resulting integrl cn be compute In generl, given ny y = f(x), we cn think of this s the vector function r(t) = t,f(t), Then r (t) = 1,f (t), n r (t) = 1+(f ) 2 The length of the curve y = f(x) between n b is thus b 1+(f (x)) 2 x Unfortuntely, such integrls re often impossible to o exctly n must be pproximte One useful ppliction of rc length is the rc length prmeteriztion A vector function r(t) gives the position of point in terms of the prmeter t, which is often time, but nee not be Suppose s is the istnce long the curve from some fixe strting point; if we use s for the vrible, we get r(s), the position in spce in terms of istnce long the curve We might still imgine tht the curve represents the position of moving object; now we get the position of the object s function of how fr the object hs trvele EXAMPLE 1333 Suppose r(t) = cost,sint, We know tht this curve is circle of rius 1 While t might represent time, it cn lso in this cse represent the usul ngle between the positive x-xis n r(t) The istnce long the circle from (1,,) to (cost,sint,) is lso t this is the efinition of rin mesure Thus, in this cse s = t n r(s) = coss,sins,

4 133 Arc length n curvture 341 EXAMPLE 1334 Suppose r(t) = cost,sint,t We know tht this curve is helix The istnce long the helix from (1,,) to (cost,sint,t) is s = r (u) u = cos 2 u+sin 2 u+1u = 2u = 2t Thus, the vlue of t tht gets us istnce s long the helix is t = s/ 2, n so the sme curve is given by ˆr(s) = cos(s/ 2),sin(s/ 2),s/ 2 In generl, if we hve vector function r(t), to convert it to vector function in terms of rc length we compute s = r (u) u = f(t), solves = f(t)fort, gettingt = g(s),nsubstitutethisbck intor(t)togetˆr(s) = r(g(s)) Suppose tht t is time By the Funmentl Theorem of Clculus, if we strt with rc length s(t) = r (u) u n tke the erivtive, we get s (t) = r (t) Here s (t) is the rte t which the rc length is chnging, n we hve seen tht r (t) is the spee of moving object; these re of course the sme Suppose tht r(s) is given in terms of rc length; wht is r (s)? It is the rte t which rc length is chnging reltive to rc length; it must be 1! In the cse of the helix, for exmple, the rc length prmeteriztion is cos(s/ 2),sin(s/ 2),s/ 2, the erivtive is sin(s/ 2)/ 2,cos(s/ 2)/ 2,1/ 2, n the length of this is sin 2 (s/ 2) 2 + cos2 (s/ 2) = = 1 So in generl, r is unit tngent vector Given curve r(t), we woul like to be ble to mesure, t vrious points, how shrply curve it is Clerly this is relte to how fst tngent vector is chnging irection, so first guess might be tht we cn mesure curvture with r (t) A little thought shows tht this is flwe; if we think of t s time, for exmple, we coul be trcing out the curve more or less quickly s time psses The secon erivtive r (t) incorportes this notion of time, so it epens not simply on the geometric properties of the curve but on how quickly we move long the curve 342 Chpter 13 Vector Functions EXAMPLE 1335 Consier r(t) = cost,sint, n s(t) = cos2t,sin2t, Both of these vector functions represent the unit circle in the x-y plne, but if t is interprete s time, the secon escribes n object moving twice s fst s the first Computing the secon erivtives, we fin r (t) = 1, s (t) = 4 To remove the epenence on time, we use the rc length prmeteriztion If curve is given by r(s), then the first erivtive r (s) is unit vector, tht is, r (s) = T(s) We now compute the secon erivtive r (s) = T (s) n use T (s) s the officil mesure of curvture, usully enote κ EXAMPLE 1336 We hve seen tht the rc length prmeteriztion of prticulr helix is r(s) = cos(s/ 2),sin(s/ 2),s/ 2 Computing the secon erivtive gives r (s) = cos(s/ 2)/2, sin(s/ 2)/2, with length 1/2 Wht if we re given curve s vector function r(t), where t is not rc length? We hve seen tht rc length cn be ifficult to compute; fortuntely, we o not nee to convert to the rc length prmeteriztion to compute curvture Inste, let us imgine tht we hve one this, so we hve foun t = g(s) n then forme ˆr(s) = r(g(s)) The first erivtive ˆr (s) is unit tngent vector, so it is the sme s the unit tngent vector T(t) = T(g(s)) Tking the erivtive of this we get The curvture is the length of this vector: s T(g(s)) = T (g(s))g (s) = T (t) t s κ = T (t) t s = T (t) s/t = T (t) r (t) (Recll tht we hve seen tht s/t = r (t) ) Thus we cn compute the curvture by computing only erivtives with respect to t; we o not nee to o the conversion to rc length EXAMPLE 1337 Returning to the helix, suppose we strt with the prmeteriztion r(t) = cost,sint,t Then r (t) = sint,cost,1, r (t) = 2, n T(t) = sint,cost,1 / 2 Then T (t) = cost, sint, / 2 n T (t) = 1/ 2 Finlly, κ = 1/ 2/ 2 = 1/2, s before EXAMPLE 1338 Consier this circle of rius : r(t) = cost, sint, 1 Then r (t) = sint,cost,, r (t) =, n T(t) = sint,cost, / Now T (t) = cost, sint, / n T (t) = 1 Finlly, κ = 1/: the curvture of circle is 133 Arc length n curvture 343 everywhere the inverse of the rius It is sometimes useful to think of curvture s escribing wht circle curve most resembles t point The curvture of the helix in the previous exmple is 1/2; this mens tht smll piece of the helix looks very much like circle of rius 2, s shown in figure 1331 Figure 1331 A circle with the sme curvture s the helix (AP) EXAMPLE 1339 Consier r(t) = cost,sint,cos2t, sshowninfigure1324 r (t) = sint,cost, 2sin(2t) n r (t) =, so T(t) = sint, cost, 2 sin2t Computing the erivtive of this n then the length of the resulting vector is possible but unplesnt Fortuntely, there is n lternte formul for the curvture tht is often simpler thn the one we hve: κ = r (t) r (t) r (t) 3 EXAMPLE 1331 Returning to the previous exmple, we compute the secon erivtive r (t) = cost, sint, 4cos(2t) Then the cross prouct r (t) r (t) is 4costcos2t 2sintsin2t,2costsin2t 4sintcos2t,1 344 Chpter 13 Vector Functions Computing the length of this vector n iviing by r (t) 3 is still bit teious With the i of computer we get 48cos4 t 48cos κ = 2 t+17 ( 16cos 4 t+16cos 2 t+1) 3/2 Grphing this we get 4 2 π 2 π 3π 2 2π Compre this to figure 1324 you my wnt to lo the Jv pplet there so tht you cn see it from ifferent ngles The highest curvture occurs where the curve hs its highest n lowest points, n inee in the picture these pper to be the most shrply curve portions of the curve, while the curve is lmost stright line miwy between those points Let s see why this lternte formul is correct Strting with the efinition of T, r = r T so by the prouct rule r = r T+ r T Then by Theorem 1241 the cross prouct is r r = r T r T+ r T r T = r r (T T)+ r 2 (T T ) = r 2 (T T ) becuse T T =, since T is prllel to itself Then r r = r 2 T T = r 2 T T sinθ = r 2 T using exercise 8 in section 132 to see tht θ = π/2 Diviing both sies by r 3 then gives the esire formul We use the fct here tht T is perpeniculr to T; the vector N = T / T is thus unit vector perpeniculr to T, clle the unit norml to the curve Occsionlly of use is the unit binorml B = T N, unit vector perpeniculr to both T n N

5 Exercises Fin the length of 3cost,2t,3sint, t [,2π] 2 Fin the length of t 2,2,t 3, t [,1] 3 Fin the length of t 2,sint,cost, t [,1] 4 Fin the length of the curve y = x 3/2, x [1,9] 134 Motion long curve Set up n integrl to compute the length of cost,sint,e t, t [,5] (It is teious but not too ifficult to compute this integrl) 6 Fin the curvture of t,t 2,t 7 Fin the curvture of t,t 2,t 2 8 Fin the curvture of t,t 2,t 3 9 Fin the curvture of y = x 4 t (1,1) ½ º ÅÓØ ÓÒ ÐÓÒ ÙÖÚ We hve lrey seen tht if t is time n n object s loction is given by r(t), then the erivtive r (t) is the velocity vector v(t) Just s v(t) is vector escribing how r(t) chnges, so is v (t) vector escribing how v(t) chnges, nmely, (t) = v (t) = r (t) is the ccelertion vector EXAMPLE 1341 Suppose r(t) = cost,sint,1 Then v(t) = sint,cost, n (t) = cost, sint, This escribes the motion of n object trveling on circle of rius 1, with constnt z coorinte 1 The velocity vector is of course tngent to the curve; note tht v =, so v n re perpeniculr In fct, it is not hr to see tht points from the loction of the object to the center of the circulr pth t (,,1) Recll tht the unit tngent vector is given by T(t) = v(t)/ v(t), so v = v T If we tke the erivtive of both sies of this eqution we get = v T+ v T (1341) Also recll the efinition of the curvture, κ = T / v, or T = κ v Finlly, recll tht we efine the unit norml vector s N = T / T, so T = T N = κ v N Substituting into eqution 1341 we get = v T+κ v 2 N (1342) The quntity v(t) is the spee of the object, often written s v(t); v(t) is the rte t which the spee is chnging, or the sclr ccelertion of the object, (t) Rewriting eqution 1342 with these gives us = T+κv 2 N = TT+ NN; T is the tngentil component of ccelertion n N is the norml component of ccelertion We hve lrey seen tht T mesures how the spee is chnging; if 346 Chpter 13 Vector Functions you re riing in vehicle with lrge T you will feel force pulling you into your set The other component, N, mesures how shrply your irection is chnging with respect to time So it nturlly is relte to how shrply the pth is curve, mesure by κ, n lso to how fst you re going Becuse N inclues v 2, note tht the effect of spee is mgnifie; oubling your spee roun curve quruples the vlue of N You feel the effect of this s force pushing you towr the outsie of the curve, the centrifugl force In prctice, if wnt N we woul use the formul for κ: To compute T we cn project onto v: N = κ v 2 = r r r 2 = r r r 3 r T = v v = r r r EXAMPLE 1342 Suppose r = t,t 2,t 3 Compute v,, T, n N Tking erivtives we get v = 1,2t,3t 2 n =,2,6t Then Exercises 134 4t+18t 3 T = n N = 1+4t2 +9t 4 1 Let r = cost,sint,t Compute v,, T, n N 2 Let r = cost,sint,t 2 Compute v,, T, n N 3 Let r = cost,sint,e t Compute v,, T, n N 4 Let r = e t,sint,e t Compute v,, T, n N 4+36t2 +36t 4 1+4t2 +9t 4 5 Suppose n object moves so tht its ccelertion is given by = 3cost, 2sint, At time t = the object is t (3,,) n its velocity vector is,2, Fin v(t) n r(t) for the object 6 Suppose n object moves so tht its ccelertion is given by = 3cost, 2sint, At time t = the object is t (3,,) n its velocity vector is,21, Fin v(t) n r(t) for the object 7 Suppose n object moves so tht its ccelertion is given by = 3cost, 2sint, At time t = the object is t (3,,) n its velocity vector is,2,1 Fin v(t) n r(t) for the object 8 Suppose n object moves so tht its ccelertion is given by = 3cost, 2sint, At time t = the object is t (3,,) n its velocity vector is,21,1 Fin v(t) n r(t) for the object 134 Motion long curve Describe sitution in which the norml component of ccelertion is n the tngentil component of ccelertion is non-zero Is it possible for the tngentil component of ccelertion to be while the norml component of ccelertion is non-zero? Explin Finlly, is it possible for n object to move (not be sttionry) so tht both the tngentil n norml components of ccelertion re? Explin