Curves. Differential Geometry Lia Vas

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1 Differentil Geometry Li Vs Curves Differentil Geometry Introduction. Differentil geometry is mthemticl discipline tht uses methods of multivrible clculus nd liner lgebr to study problems in geometry. In this course, we will study curves nd surfces nd, lter in the course, their generliztions mnifolds. A pivotl concept tht we shll develop both for curves nd for surfces is tht of curvture. The long term gol is to provide the mthemticl bckground necessry for understnding Einstein s theory of reltivity s geometric theory of spce-time in which the grvittion mnifests s the curvture of spce-time. We begin by studying curves nd their properties such s velocity nd ccelertion vectors, curvture nd torsion nd rc length. Then we introduce n pprtus tht completely describes ll the relevnt properties of curve. After tht, we move on to surfces nd their properties. We shll study the first nd second fundmentl forms, geodesics, nd curvture n pprtus tht completely describes surfce. We shll formulte Guss s Theorem Egregium (Remrkble Theorem) tht llows the concept of curvture to be generlized to curvture of higher dimensionl mnifolds nd enbles you to understnd the lnguge used in specil nd generl reltivity. nd to generlize the content of this course to higher dimensions. Curves. Intuitively, we think of curve s pth trced by moving prticle in spce. This pproch is formlized by considering curve s function of prmeter, sy t. Thus, the domin of curve is n intervl (, b) (possibly (, )) consisting of ll possible vlues of prmeter t. The rnge of curve is contined in the three dimensionl spce. To simplify the nottion, consider tht R is frequently used to denote the set of rel numbers. If we think of points in the two dimensionl spce s the set of ordered pirs (x, y) where ech coordinte is in R, the product of R with itself is considered nd is denoted by R. Following this resoning, the three-dimensionl spce is denoted by R 3. Thus, curve γ is mpping of n intervl (, b) into R 3. This is denoted by γ : (, b) R 3. Using the stndrd vector representtions of points in R 3 s (x, y, z), we cn represent γ s vector function: γ(t) = (x(t), y(t), z(t)) or using the prmetric equtions x = x(t), y = y(t), nd z = z(t). The vrible t is clled the prmeter. 1

2 Exmples. 1. Line. A line in spce is given by the equtions x = x 0 + t y = y 0 + bt z = z 0 + ct where (x 0, y 0, z 0 ) is point on the line nd (, b, c) is vector prllel to it. Note tht in the vector form the eqution γ = γ(0) + m t for γ(0) = (x 0, y 0, z 0 ) nd m = (, b, c), hs exctly the sme form s the well known y = b + mx.. Circle in horizontl plne. Consider the prmetric equtions x = cos t y = sin t z = b. Recll tht the prmetric eqution of circle of rdius centered in the origin of the plne R re x = cos t, y = sin t. Recll lso tht z = b represents the horizontl plne pssing b in the z-xis. Thus, the equtions x = cos t y = sin t z = b represent the circle of rdius in the horizontl plne pssing z = b on z-xis. 3. Ellipse in plne. Consider the intersection of cylinder nd plne. The intersection is n ellipse. For exmple, if we consider cylinder with circulr bse x = cos t, y = sin t nd the eqution of the plne is mx + ny + kz = l with k 0, the prmetric equtions of ellipse cn be obtined by solving the eqution of plne for z nd using the equtions for x nd y to obtin the eqution of z in prmetric form. Thus z = 1(l mx ny) k nd so x = cos t y = sin t z = 1 (l m cos t n sin t). k 4. Circulr helix. A curve with equtions x = cos t y = sin t z = bt is the curve spirling round the cylinder with bse circle x = cos t, y = sin t. 5. Plne curves. All the concepts we develop for spce curves correspond to plne curves simply considering tht z = 0.

3 Tngent Vector nd the Arc Length Velocity vector. If γ(t) = (x(t), y(t), z(t)) is spce curve, the vector tngent to the curve t the point where t = t 0 is γ (t 0 ) = (x (t 0 ), y (t 0 ), z (t 0 )). So, the eqution of the line tngent to the curve γ t the point where t = t 0 is or, using the shorter nottion, x = x(t 0 ) + x (t 0 )t y = y(t 0 ) + y (t 0 )t z = z(t 0 ) + z (t 0 )t The derivtive vector γ (t) = (x (t), y (t), z (t)) is sometimes lso clled the velocity vector. The length of this vector is clled the speed, sometimes lso referred to s instntneous speed nd denoted by ds dt so ds dt = γ (t) = (x (t)) + (y (t)) + (z (t)). Also note tht ds = γ (t) dt. γ(t 0 ) + γ (t 0 )t. Derivtive vector of unit length is especilly importnt. It clled the tngent vector T(t) nd is obtined by dividing the derivtive vector by its length T(t) = 1 γ (t) (x (t), y (t), z (t)) = γ (t) γ (t). The totl length of the curve γ on intervl t b cn be obtined by integrting the length element ds from to b. Thus, b b b L = ds = γ (t) dt = (x (t)) + (y (t)) + (z (t)) dt Reprmetriztion. For one-to-one mpping of n intervl (c, d) onto intervl (, b) given by s t(s), the curve γ(t) = (x(t), y(t), z(t)) cn be given by γ(s) = (x(t(s)), y(t(s)), z(t(s))). In this cse, γ(s) is sid to be reprmetriztion of γ(t). A reprmetriztion does not chnge the grph of curve - it cn be considered to chnge just the speed of the curve moving long the curve. 3

4 Exmple 1. The line with prmetric equtions x = 1 + t y = + 4t z = 3 t cn lso be represented s x = 1 + s y = + s z = 3 s with s = t being the reltion between the two prmeters just s in the figure bove. Further, it x = 3 u cn lso be represented s y = 6 4u z = 1 + u with u = 1 t being the reltion between the prmeters in the first nd the lst prmetriztion, gin just s in the figure bove. This shows tht seemingly different prmetric equtions cn be describing the sme line. When thinking of line s the trjectory of n object nd the prmeter s the time, the different prmetriztion represent the fct tht different prticles trveling on the sme (stright) pth but with different speeds nd possibly in the opposite orienttion re still moving long the sme pth. With this in mind, the prticle in prmetriztion prticle in prmetriztion x = 1 + s y = + s z = 3 s. x = 1 + t y = + 4t z = 3 t trvels twice s fst thn the A third prticle with the trjectory given by x = 3 u y = 6 4u z = 1 + u hs the different initil point nd it is trveling in different orienttion thn the first two but ll three prticles re trveling long the sme pth. Exmple. Consider the cycle centered t the origin of rdius. These prmetric equtions x = cos t nd y = sin t eqully well represent the circle s the usul x = cos t nd y = sin t. The difference is tht it tkes point 4π to mke one full rottion using the first prmetriztion while it tkes just π to mke one full rottion using the second prmetriztion. So, the point moves twice s slow in the first representtion thn in the second. The sme conclusion cn be obtined when compring speeds (lengths of the velocity vectors) in both representtions. In the first one, the velocity vector is ( sin t, cos t ) nd its length is while in the second the velocity vector is ( sin t, cos t) nd its length is, twice s lrge speed thn in the first prmetriztion. More generlly, the prmetriztion x = cos t nd y = sin t where b is positive constnt, b b hs the velocity ( sin t, cos t) nd speed. By tking b =, we obtin the prmetriztion of b b b b b unit speed. x = cos t nd y = sin t One would prefer to be ble to obtin the prmetriztion with unit speed for every curve γ. In such prmetriztion, the velocity vector hs unit length nd so γ = T. We refer to curve prmetrized in tht wy s unit-speed curve. The unit-speed prmetriztion cn be obtined s reprmetriztion of ny given prmetriztion using the substitution t = t(s) such tht s(t) = t ds = t γ (t) dt = t (x (t)) + (y (t)) + (z (t)) dt This prmetriztion is lso referred s the prmetriztion by the rc length nd s is used 4

5 to denote the prmeter in this cse. Thus, T(s) = γ (s) nd γ (s) = 1 If the curve is prmetrized by rc length, the length L for t b which corresponds to c s d in the rc-length prmetriztion, cn lso be found s L = b γ (t) dt = d c γ (s) ds = d c 1ds = (d c). Exmple 1. Consider the line (1 + t, + t, 3 t) from the erlier exmple. To prmetrize it by the rc length, we clculte the prmetriztion s = s(t) to be t dt = 6t. Thus, t = 1 6 s nd we hve ( s, + 6 s, s). Note lso tht ( ) is the normliztion of the originl velocity vector (1,, 1). Exmple. We cn verify our erlier conclusion tht the substitution t = s gives the unit-speed prmetriztion of the circle x = cos t, y = sin t by clculting tht s = t dt = t. Thus 0 t = s nd the unit-speed representtion is x = cos s, y = sin s. Note tht unit-speed prmetriztion is not lwys possible in some cses the ntiderivtive (x (t)) + (y (t)) + (z (t)) dt cnnot be found s n elementry function. In some other cses, it my not be possible to solve s = s(t) = t (x (t)) + (y (t)) + (z (t)) dt for t. A clss of curves tht lwys llows the prmetriztion by rc length includes the curves we consider next. Regulr curves. A curve γ(t) = (x(t), y(t), z(t)) is sid to be regulr if the functions x(t), y(t), nd z(t) re continuous with continuous derivtives nd such tht the tngent vector (x (t), y (t), z (t)) is not zero. Considering regulr curves rules out some pthologicl cses. In prticulr, Continuity requirement gurntees no holes or jumps. Continuity of the derivtive gurntees no corners or shrp turns (consider the grph of bsolute vlue function for exmple, or y = x /3 ) This condition gurntees tht the tngent vector of unit length cn be found t every point. This condition lso implies tht the reprmetriztion by rc length will be possible. The sme curve cn hve regulr nd non-regulr prmetriztion. For exmple, the x-xis cn be given s (t, 0, 0) but lso s (t 3, 0, 0). The first prmetriztion is regulr while the second is not since the derivtive (3t, 0, 0) is 0 d t = 0. Prctice problems. 1. Find n eqution of the line through the point (, 4, 10) nd prllel to the vector (3, 1, 8). Check if (4, 6, 6) nd (1, 4, 4) re on the line.. Find n eqution of the line through the points (3, 1, 1) nd (3,, 6). 5

6 3. Find the eqution of tngent line t the point where t = 0. Find lso unit tngent vector T s function of prmeter t. () Line x = 1 + t y = t z = 1 + t. (b) Circle in horizontl plne x = cos t y = sin t z =. (c) Circulr helix x = cos t y = sin t z = t. (d) Ellipse in the intersection of the cylinder x + y = 1 with the plne y + z = (find prmetric equtions before finding the tngent line). 4. Consider the curve C which is the intersection of the surfces x + y = 9 nd z = 1 y. i) Find the prmetric equtions tht represent the curve C. ii) Find the eqution of the tngent line to the curve C t point (0, 3, 8). 5. Consider the curve C which is the intersection of the surfces y + z = 16 nd x = 8 y z i) Find the prmetric equtions tht represent the curve C. ii) Find the eqution of the tngent line to the curve C t point ( 8, 4, 0). 6. Consider the helix x = cos t, y = sin t, z = t. () Find the rc length of the helix for 0 t π using the given prmetriztion. (b) Prmetrize the helix by the rc length. (c) Find the rc length of the helix for 0 t π using the rc-length prmetriztion. 7. Prmetrize the curve x = e t cos t, y = e t sin t, z = e t by the rc length. Solutions. 1. x = +3t y = 4+t z = 10 8t. Yes. No..x = 3 y = 1+t z = 1 5t (or x = 3 y = + t z = 6 5t). 3. () Line; the tngent line is the sme s the line itself. T = 1(1,, ) = ( 1,, ) (b) Circle; the tngent line is x = 1 y = t z =. T = ( sin t, cos t, 0) (c) Helix; the tngent line is x = 1 y = t z = t. T = 1 ( sin t, cos t, 1) (d) Ellipse, the prmetric equtions re x = cos t, y = sin t, z = sin t. The tngent line is x = 1 y = t z = t. 1 T = 1+cos ( sin t, cos t, cos t) t 4. i) x = 3 cos t, y = 3 sin t, z = 1 y = 1 9 sin t. ii) (0, 3, 8) corresponds to t = π/. Plugging π/ in derivtive gives you ( 3, 0, 0). Tngent line: x = 3t y = 3 z = i) y = 4 cos t, z = 4 sin t, x = 8 y z = 8 16 cos t 4 sin t. ii) ( 8, 4, 0) corresponds to t = π. Plugging π in derivtive gives you (4, 0, 4). Tngent line: x = 4t 8 y = 4 z = 4t. 6. () L = π γ dt = π 0 0 ( sin t) + ( cos t) + dt = π dt = π dt = π. 0 0 (b) s = t + 0 dt = t t = s. Thus the rc-length prmetriztion is x = cos s, y = sin s, z = s. (c) First note tht when t = 0, s = 0 s well nd when t = π, then s = π. Thus the bounds for s re 0 nd π. So, using the rc-length prmetriztion L = π γ (s) ds = 0 π 1 ds = π. Since we got the sme nswer s in prt ), we cn hypothesize tht the length 0 of curve should not be dependent on specific prmetriztion. 7. s = t 0 e t + e t dt = 3e t (s + 3) = e t t = ln( 1 3 (s + 3)). Thus the rclength prmetriztion is obtined by substituting ln( 1 3 (s + 3)) for t in the given prmetriztion. 6

7 Accelertion Vector nd Curvture The derivtive of velocity vector is clled ccelertion vector γ (t) = (x (t), y (t), z (t)). This vector hs the sme direction s the force needed to keep the prticle on the trck of γ (i.e. mke the prticle trverse the curve). This force mkes prticle trveling long γ to sty on this course. Without this force, such prticle would continue the motion s indicted by the velocity vector nd not sty on the course of γ. Exmple. The ccelertion of the circle x = cos t, y = sin t is x = cos t, y = sin t so γ (t) = γ(t) in this cse. Thus, the ccelertion is directed towrds the center nd of constnt mgnitude. Consider the specil cse when the curve is prmetrized with respect to rc length s. In this cse, the length of the ccelertion vector is clled the curvture κ(s) = T (s) = γ (s). Thus, the curvture mesures the intensity of the force needed to keep the prticle on the trck of γ. This formul lso illustrtes tht the curvture mesures how quickly the curve chnges direction since it mesures the rte the of chnge of direction vector s the vlue of prmeter s chnges. Exmple 1. To find the curvture of line γ = (x 0 + t, y 0 + bt, z 0 + ct), compute tht s = t 0 γ (t) dt = + b + c t so tht T = γ 1 (s) = (, b, c). Since this is constnt vector, +b +c T = (0, 0, 0) nd so κ = T = 0. Thus, the curvture of line is 0. This grees with the intuitive ide of the curvture: stright line does not curve t ll. Exmple. To find the curvture of the circle γ = ( cos t, sin t), recll tht the unit-length prmetriztion is given by t = s so tht γ = ( cos s, sin s ). Compute tht T = γ = ( sin s, cos s) nd so T = γ = ( 1 cos s, 1 sin s ) which gives us κ = T 1 = cos s + 1 sin s = 1 = 1. Thus, circle of smll rdius hs lrge curvture nd circle of lrge rdius hs smll curvture. 7

8 This exmple lso grees with the intuitive ide of the curvture: the circle curves eqully t every point. So, the curvture of circle is constnt. These two exmples generlize so tht firly stright curve hs the direction of derivtive vector chnging very slowly, so tht curvture is smll. If curve bends or twists shrply, the direction of derivtive vector chnges quickly. The curvture in non-unit-speed prmetriztions. When the unit-speed prmetriztion leds to complex formuls, it my be esier to compute the curvture using ny given prmetriztion. The formul for computing κ is κ = γ γ γ 3 nd cn be obtined s follows. 1. Strt by observing tht γ (t) nd T(s) re relted by γ (t) = dγ dt = dγ ds ds dt = Tds dt.. Find derivtive of the eqution from (1) with respect to t. γ = dt ds dt dt + s Td dt = dt ds 3. Cross multiply this eqution by γ = T ds, to get dt Since T T = 0, ds ds dt dt + s Td dt = T ( ds dt ) + T d s dt. γ γ = T T( ds dt )3 + T T d s dt ds dt. γ γ = T T( ds dt )3. 4. Consider the length of both sides of the lst eqution nd get γ γ = T T ds dt 3. Now note tht the fct tht length of T is 1 implies tht 1 = T = T T T T = 1. Differentiting T T = 1 we hve tht T T + T T = 0 T T = 0 T T = 0. Thus T nd T re orthogonl. So the length of the cross product T T cn be computed s T T sin( π ) which is equl to T = κ since T = 1. Also note tht ds = dt γ. Thus we hve tht γ γ = κ γ 3. Solving for κ we obtin the formul for the curvture to be κ = γ γ. γ 3 8

9 Norml nd Binorml Vectors. Torsion The unit vector in the direction of T is clled the principl norml vector nd is denoted by N. Thus, N(s) = T (s) = T (s) T (s) κ(s) Recll tht T nd T re orthogonl nd so N nd T re orthogonl. The plne determined by vectors T nd N is clled the osculting plne. If curve is plnr, this plne is the plne in which curve lies. The nme osculting comes from the following construction. At point on the curve, consider the circle of rdius 1 with the sme tngent s the curve nd with κ the center on the norml direction line. Thus, it lies in the osculting plne. This circle is clled the osculting circle. The nme comes from Ltin word osculri mens to kiss (osculum mening kiss) nd refers to the fct tht curve nd osculting circle re tngentil curves. The osculting circle is n pproximtion of the curve with circle t point - it hs the sme curvture, tngent nd norml vector s the curve t point. The center of the circle cn be computed by the formul γ + 1 κ N. To complete n pprtus tht fully describes the spce curves, we need one more vector besides T nd N (this should not be surprising considering tht there should be three bsis vectors). Consider the cross product of T nd N. B = T N This vector is clled the binorml vector. By definition, B is perpendiculr both to T nd N nd hs unit length lso since T nd N re of unit length. The plne determined by N nd B is clled the norml plne. All the lines in the norml plne re perpendiculr to the tngent vector. 9

10 Vector B indictes the direction in which the curve deprts from being plnr curve. The triple T, N, B is cll the moving frme. Wtch the nimtion on Wikipedi (do serch for FrenetSerret formuls ) to understnd the moving frme better. The moving frme T, N, B is n orthonorml bsis mening tht ech vector is of unit length, tht ech two re perpendiculr nd tht every other vector cn be uniquely represented s liner combintion of these three. This mens tht they tke over the role of the usul bsis vectors i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1), t point on the curve. The stndrd bsis my be unfitted for describing geometry of the curve. Thus, the chnge of bsis is needed to obtin geometrics bsis fitted for the curve. Torsion. While B indictes the direction in which the curve deprts from being plnr curve, the torsion τ mesures the extent t which the curve deprts from being plne curve, tht is, it mesures how much the curve deprts from the norml plne. Torsion τ cn be computed vi N nd B s follows. Recll tht B is defined s T N. Differentiting with respect to s we obtin B = T N+T N. Substituting tht T = κn nd noting tht N N = 0, we obtin tht B = 0 + T N. Thus, B is orthogonl to both T nd N. Since N is lso orthogonl to T nd N (differentiting N N = 1 obtin tht N N = 0), we conclude tht B nd N re co-liner nd so B is multiple of N. The proportionlity constnt is denoted by τ(s) nd so B = τn Let us (dot) multiply ech side of the eqution B = τn by N. This gives us tht the torsion τ cn be computed s τ = B N For plne curve, the curve lies in the osculting plne, thus τ = 0. If τ > 0, the curve twists towrds the side tht B points to. If τ < 0, the curve twists wy from the side tht B points to. Exmple. Reprmetrizing by rc length first, compute the curvture, torsion nd the moving frme of the helix ( cos t, sin t, t). Solution. The rc-length prmetriztion of the helix is given by t = s (see problem 6 of previous section.) Thus γ(s) = ( cos s, sin s s, ) so tht T = γ 1 = ( sin s, cos s, 1) nd T = 1 1 ( cos s, sin s, 0). 10

11 So κ = T = 1. Thus, the helix hs constnt curvture of 1. N = T = ( cos T s, sin s, 0) nd B = T N = 1 (sin s, cos From here we find B = 1 1 (cos s, sin s, 0) = 1 (cos s, sin Thus the torsion is lso constnt. s, 1). s, 0) nd τ = B N = 1. Non-unit-speed prmetriztion formuls. Sometimes, it is esier to use the formuls for the moving frme nd curvture tht re not given in terms of rc length prmetriztion. We hve seen tht T = γ in this cse nd hve derived the formul for the curvture to be κ = γ γ. γ γ 3 The formuls for N, B nd τ given below complete the description of the curvture, torsion nd the moving frme for ny prmetriztion. T = γ, κ = γ γ, B = γ γ, N = B T, nd τ = (γ γ ) γ γ γ 3 γ γ γ γ The formul for τ cn be shortened using the following nottion. Let [, b, c] denotes the vlue of 3 3 determinnt with columns, b, c. Then [, b, c] = ( b) c nd so τ = [γ,γ,γ ]. γ γ Exmple. Using the given prmetriztion, compute the curvture, torsion nd the moving frme of the helix ( cos t, sin t, t). Solution. Here γ = ( sin t, cos t, ), γ = =, γ = ( cos t, sin t, 0) nd γ γ = (sin t, cos t, 1). Thus γ γ =. This gives us T = 1 ( sin t, cos t, 1) nd κ = ( = 1 tht grees to wht we clculted before. ) 3 3 Then clculte tht B = 1 (sin t, cos t, 1) = 1 (sin t, cos t, 1), N = B T = ( cos t, sin t, 0) nd τ = 1 tht lso grees wht we clculted before. Frenet-Serret pprtus The moving frme T, N, B together with curvture κ nd torsion τ provide complete list of geometricl properties of the curve. The three vector functions T, N, B nd two sclr functions κ nd τ re known s Frenet-Serret pprtus. The fundmentl theorem of differentil geometry of curves sttes tht curve is completely determined by its curvture nd torsion lone. Specificlly, this mens tht the moving frme T, N, B is completely determined by the two functions κ nd τ. This sttement is known s Frenet- Serret theorem. It follows from two fcts. 1. The derivtives T, N, B cn be represented s T = κn N = κt +τb B = τn The fist equtions follows from the definition of the norml vector N nd the third follows from the definition of torsion τ. To obtin the second eqution, note tht the orthonorml system T, N, B is such tht T N = B, N B = T, nd B T = N 11

12 Differentiting the lst formul, we hve N = B T+B T = τn T+B κn = τb κt which is the second eqution. This system of first order differentil equtions is known s Frenet-Serret system.. The theory of first order differentil equtions gurntees tht the bove system hs unique solution for ny vlue of initil conditions nd ny vlues of κ nd τ. The Frenet-Serret formuls were first obtined by Frenet in 1847 in his disserttion. Serret first published them in Their importnce lies in their pplictions in severl different res. In kinemtics, the Frenet-Serret formuls describe the kinemtic properties of prticle which moves long curve. In the life sciences, prticulrly in models of microbil motion, considertions of the Frenet-Serret frme hve been used to explin the mechnism by which moving orgnism in viscous medium chnges its direction. In reltivity theory clssicl coordinte system is not convenient for trjectory. In cses like tht the Frenet-Serret frme is used. Within this setting, Frenet-Serret frmes hve been used to model the precession of gyroscope in grvittionl well. In erodynmics, ircrft mneuvers cn be expressed in terms of the moving frme. Other res of pplictions include mterils science, elsticity theory, nd computer grphics. The Frenet-Serret theorem hs the following corollries. A curve with zero curvture nd torsion is stright line. (Moreover, it cn be shown tht curve with constnt curvture is stright line). A curve with zero torsion is plnr curve. A curve with constnt curvture nd zero torsion is prt of circle. A curve with constnt curvture nd constnt torsion is helix. To illustrte the methods for proving these corollries, we present proof of the following corollry. Corollry. If γ is regulr curve, the following conditions re equivlent. 1. γ is plnr curve (i.e. there is plne tht contins γ).. The torsion τ is identiclly zero. 3. The binorml vector B is constnt. If you hve not encountered mthemticl proofs before, consider the footnote below before going over the proof of this corollry. 1 1 Some methods of proving mthemticl sttements. Mny mthemticl sttements we encounter in this course re formulted s implictions or equivlences. (1) Impliction. An impliction is n expression of the form if p, then q (equivlently, p implies q or q follows from p). In symbols, this is expressed s p q. To prove n impliction of this form, you cn ssume tht p holds nd deduce tht q holds in this cse s well. Another wy to prove n impliction p implies q is to use its contrpositive not p implies not q. For exmple, contrpositive of the sttement If polygon hs three sides, it is tringle. is If polygon is not tringle, then it does not hve three sides. Yet nother wy to prove n impliction of the form p implies q is to show tht simultneously ssuming p nd 1

13 Proof of Corollry. We shll show tht the three conditions re equivlent by showing tht (1) implies (), () implies (3) nd (3) implies (1). Assume tht γ lies in the plne x + by + cz = d. Then this plne is the osculting plne of the curve γ. So the vector B is orthogonl to this plne t every point on γ. Since v = (, b, c) is vector perpendiculr to the plne, the vector B t every point is either 1 1 v or v. In both cses, B is v v constnt vector. Thus B = 0 nd so τ = B N = 0. Thus, the torsion is identiclly zero. This proves tht (1) implies (). Assume now tht () holds nd show (3). If τ is identiclly zero, then B = τn is lso identiclly zero. Thus B is constnt vector. Finlly, let us show tht (3) implies (1). Recll tht n eqution describing ny point x = (x, y, z) of plne pssing x 0 = (x 0, y 0, z 0 ) perpendiculr to vector v = (, b, c) is (x x 0 ) v = 0. So, to show tht γ lies in plne, it is sufficient to show tht the dot product (γ(s) γ(0)) B = 0. Consider the derivtive of this dot product ((γ(s) γ(0)) B) = γ (s) B + (γ(s) γ(0)) B The second term (γ(s) γ(0)) B is zero since B = 0. The vector γ (s) is T nd T is perpendiculr to B so the first term γ (s) B is zero s well. Hence (γ(s) γ(0)) B is constnt for every s. But since for s = 0 this expression is 0, this constnt hs to be zero for every s. So, (γ(s) γ(0)) B = 0. This shows tht γ lies in the plne pssing γ(0) tht is perpendiculr to B. So (3) implies (1). Prctice Problems 1. Find the curvture of the twisted cubic (t, t, t 3 ).. Find the formul clculting the curvture of the prbol y = x in terms of x. Use it to clculte the curvture t x = 0 nd x = Find the osculting circle for prbol y = x t the origin. 4. Find the Frenet-Serret pprtus for the curve ( 5 curve is prmetrized by the rc length. 5. Find the osculting plne of the curve from the previous problem cos s, sin s, cos s). Note first tht the Find eqution of one of the cylindricl surfces tht intersecting the plne from (5) cretes the curve from (4). the opposite of q yields contrdiction. This form of the proof is clled reductio d bsurdum (Ltin: reduction to the bsurd ). For exmple, let us show tht if 0, n eqution of the form x = b hs only one solution. Let us ssume the opposite, tht there re two different solutions x 1 nd x. Since both solutions stisfy the eqution, we hve tht x 1 = b nd x = b. If we subtrct the second eqution from the first, we get (x 1 x ) = 0. When we divide by 0 we hve tht x 1 x = 0 so x 1 = x. But this is in contrdiction with the ssumption tht x 1 is different thn x. () Equivlence. If p implies q nd q implies p s well, we sy tht p nd q re equivlent. In symbols, this is written s p q. In this cse, p is true if nd only if q is true. A sttement of the form p if nd only if q cn be proven in two steps: (1) show tht q holds ssuming p nd, () show tht p holds ssuming q. (3) Chin of Equivlences. A sttement of the form p, q nd r re ll equivlent, cn be shown by demonstrting tht p q nd q r becuse then p r lso. Alterntively, if closed chin of implictions p q, q r, nd r p is shown to hold, then ll three conditions re equivlent. The proof of the corollry bove will illustrte this method. 13

14 7. Find the Frenet-Serret pprtus for the twisted cubic (t, t, t 3 ). Solutions. (1) γ = (1, t, 3t ), γ = 1 + 4t + 9t 4, γ = (0,, 6t), γ γ = ( 6t, 6t, ), γ γ = 36t4 + 36t + 4 = 9t 4 + 9t + 1 κ = 9t 4 +9t +1 (1+4t +9t 4 ) 3/. () Considering the prmetriztion (x, x, 0), we find γ = (1, x, 0), γ = 1 + 4x, γ = (0,, 0), γ γ = (0, 0, ). Thus, κ = (1+4x ) 3/. κ(0) =, κ(1) = (3) The rdius of the circle is = 1 by κ(0) previous problem. The center is given by γ(0) + 1 N(0). Since N is orthogonl to T nd the tngent to prbol t the origin is the x-xis, N is κ(0) prllel to y-xis nd so N = (0, 1). γ(0) = (0, 0) nd so the center is t (0, 0) + 1 (0, 1) = (0, 1 ). Hence the circle is given by ( 1 cos t, sin t). (4) The curve is unit-speed since γ = ( 5 1 sin s, cos s, sin s) nd γ = 1. Thus, T = γ. T = γ = ( 5 1 cos s, sin s, cos s) κ = T = 1, thus N = T nd B = T T = ( 1 5, 0, ) so B = 0 nd thus τ = 0. Note tht the results κ = 1 nd τ = 0 tells us tht, lthough well disguised, the curve is circle in plne. (5) From problem (4) we know tht the curve lies in plne. This plne is equl to the osculting plne nd so the osculting plne does not depend on the choice of the point on the curve. The vector B = ( 1 5, 0, ) is the norml vector of the osculting plne. For the point on the plne, we cn use ny point on the curve, for exmple the point corresponding to s = π/ (to mke the most terms zero) nd so we hve (0, , 0). We obtin the eqution of the plne (x 0) + 0(y + ) 5 (z 0) = x 5 1 z = 0 1x 5z = 0 z = x Note tht without clculting the torsion, it is impossible to tell tht the curve lies in plne simply by looking t the prmetriztion. (6) One of the simplest solutions is to consider the projections of the curve in xy plne. We obtin the ellipse x = 5 8 cos s nd y = sin s. This ellipse cn be described by x + (y (5/13) 13 ) = 1. This is n ellipse centered t (0, 8 5 ) with semixes nd (7) In problem (1) we hve found the curvture. T = 1+4t (1, t, 3t ), +9t 4 1 B = 9t 4 +9t +1 (6t 1, 6t, ) N = 1+4t +9t 4 9t 4 +9t +1 ( 18t3 4t, 18t 4, 1t 3 +6t), γ = (0, 0, 6), 1 τ = = 3. 4(9t 4 +9t +1) 9t 4 +9t +1 14

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