Surfaces. Differential Geometry Lia Vas

Transcription

1 Differentil Geometry Li Vs Surfces When studying curves, we studied how the curve twisted nd turned in spce. We now turn to surfces, two-dimensionl objects in three-dimensionl spce nd exmine how the concept of curvture trnsltes to surfces. In Clculus 3, you hve encounter surfces defined s grphs of rel vlued functions of two vribles z = f(x, y). This function lso cn tke the form x = f(y, z) or y = f(x, z). In some cses, on the other hnd, this function is given implicitly s F (x, y, z) = 0. For exmple, sphere of rdius is given by x + y + z = nd it is impossible to get single two vrible function tht would describe the whole sphere. Cylinder x + y = is nother such exmple. Let us review some exmples from Clculus 3. Plnes. The generl eqution of plne is x + by + cz + d = 0. A plne is uniquely determined by point in it nd vector perpendiculr to it. The eqution tht describes ny point x = (x, y, z) in the plne through point x 0 = (x 0, y 0, z 0 ) perpendiculr to vector = (, b, c) is (x x 0 ) = 0 The bove vector eqution of the plne hs the following sclr form. (x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0. Surfces of revolution. z = f( x + y ). To get grph of such surfce, grph the function z = f(y) in yz-plne nd let it rotte bout z-xis. For exmple A cone z = x + y is obtined by rotting the line z = y. A prboloid z = x + y is obtined by rotting the prbol z = y. A hlf-sphere z = x y is obtined by rotting the hlf-circle z = y. Cone Prboloid Hemisphere 1

2 Cylindricl surfces. These surfces re given by n eqution in which one vrible is not present. For exmple, z = f(y). To grph this surfce, grph the function z = f(y) in yz-plne nd trnslte the grph in direction of x-xis. For exmple, The grph of y + z = 4 is cylinder obtined by trnslting the circle y + z = 4 of rdius centered t the origin in yz-plne long x-xis. The grph of x + y = 4 is cylinder obtined by trnslting the circle x + y = 4 of rdius centered t the origin in xy-plne long z-xis. Cylinder x + y = 4 Cylinder y + z = 4 Prmetric Surfces In cses when surfce is given s n implicit function F (x, y, z) = 0, it my be useful to describe the three vribles x, y nd z but using some other prmeters u nd v. In tht cse, we hve tht x = x(u, v) y = y(u, v) z = z(u, v). These equtions re clled prmetric equtions of the surfce nd the surfce given vi prmetric equtions is clled prmetric surfce. Thus, prmetric surfce is represented s vector function of two vribles, i.e. the domin D consisting of ll possible vlues of prmeters u nd v is contined in R. The rnge of the surfces is contined in the three dimensionl spce R 3. Thus, surfce x is mpping of D into R 3. This is denoted by x : D R 3. The vector function x cn lso be represented s x(u, v) = (x(u, v), y(u, v), z(u, v)). Notice n nlogy with curves. We cn think of curves s one-dimensionl objects in threedimensionl spce nd surfces s two-dimensionl objects in three dimensionl spce. Thus, curve cn be described using single prmeter t. Surfce, on the other hnd, is described using two prmeters u nd v. Mpping Dimension Prmeter(s) Equtions Curve γ : (, b) R R 3 1 t γ(t) = (x(t), y(t), z(t)) Surfce x : D R R 3 u, v x(u, v) = (x(u, v), y(u, v), z(u, v))

3 The curves given in the form z = f(x, y) cn lwys be prmetrized s x = (x, y, f(x, y)). For exmple, the plne x + 3y + z = 6 cn be represented by x = (x, y, 6 x 3y). Some other surfces require fncier prmetriztions. Recll the following chnge of coordintes used in Clculus 3 tht provided prmetric representtions of some frequently used surfces. Cylindricl coordintes. x = r cos θ, y = r sin θ, z = z. Here x nd y re converted using polr coordintes nd the only chnge in z my come just from chnges in x nd y. Note tht in these coordintes. x + y = r Using cylindricl coordintes we obtin prmetriztions in the following exmples. 1. The prboloid z = x + y cn be represented s x = r cos θ, y = r sin θ, z = r, since x + y = r in cylindricl coordintes so z = r. In this course, we write this prmetriztion shortly s x = (r cos θ, r sin θ, r ). Note tht this prboloid cn lso be prmetrized by (x, y, x + y ).. The cone z = x + y cn be represented by x = (r cos θ, r sin θ, r). 3. The cylinder x + y = 4 is such tht the r vlue is constnt nd equl to. Thus, the two remining prmeters, θ nd z cn be used for representtion of this cylinder s x = cos θ, y = sin θ, z = z or, written shortly s x = ( cos θ, sin θ, z). 4. Similrly, the cylinder y + z = 4 cn be prmetrized by x = (x, cos θ, sin θ). Sphericl coordintes. If P = (x, y, z) is point in spce nd O denotes the origin, let r denotes the distnce of the point P = (x, y, z) from the origin O. Thus, x + y + z = r ; θ is the ngle between the projection of vector OP = (x, y, z) on the xy-plne nd the vector i = (1, 0, 0) (positive x xis); nd φ is the ngle between the vector OP nd the vector k = (0, 0, 1) (positive z-xis). 3

4 With this nottion, sphericl coordintes re (r, θ, φ). The conversion equtions re x = r cos θ sin φ y = r sin θ sin φ z = r cos φ. In this prmetriztion, the north pole of sphere centered t the origin corresponds to vlue φ = 0, the equtor to φ = π nd the south pole to φ = π. To mtch the geogrphicl ltitude (for which north nd south pole correspond to π π nd nd equtor to φ = 0), the ngle φ is often considered to be the ngle between the equtor nd the vector OP. In this cse, cos φ nd sin φ re switched in the equtions of the sphericl coordintes nd we obtin x = r cos θ cos φ y = r sin θ cos φ z = r sin φ. The ngle θ corresponds to the geogrphicl longitude nd the ngle φ corresponds to the geogrphicl ltitude. For exmple, the sphere x + y + z = 9 hs representtion s x = (3 cos θ sin φ, 3 sin θ sin φ, 3 cos φ) or, using the second version, s x = (3 cos θ cos φ, 3 sin θ cos φ, 3 sin φ). The upper hlf of the sphere x + y + z = 9 cn be prmetrized on severl different wys. Using x, y s prmeters x = (x, y, 9 x y ) for x + y 9. Using cylindricl coordintes x = (r cos θ, r sin θ, 9 r ) for 0 θ π, nd 0 r 9. Using sphericl coordintes x = (3 cos θ sin φ, 3 sin θ sin φ, 3 cos φ) for 0 θ π nd 0 φ π. Tngent Plne For prmetric surfce x = (x(u, v), y(u, v), z(u, v)), the derivtives x u nd x v re vectors in the tngent plne. Thus, their cross product x u x v = (x u, y u, z u ) (x v, y v, z v ) is perpendiculr to the tngent plne. If surfce is given by implicit function F (x, y, z) = 0, then this cross product lso corresponds to the grdient F of F. Prctice Problems. 1. Find n eqution of the plne through the point (6, 3, ) nd perpendiculr to the vector (, 1, 5). Check if (, 1, 0) nd (1,, 1) re in tht plne.. Sketch the following surfces. () z = 6 x 3y (b) z = 9 x y (c) z = 1 x +y (d) z = y 4

5 3. Find the eqution of the tngent plne to given surfce t the specified point. () Hyperbolic prboloid z = y x, t ( 4, 5, 9) (b) Ellipsoid x + y + 3z = 1, t (4, 1, 1). (c) Prmetric surfce given by x = u + v y = 3u z = u v, t (, 3, 0) (d) The cylinder x + z = 4, t (0, 3, ). Solutions. (1) (x 6) + 1(y 3) + 5(z ) = 0 x + y + 5z = 1. No. Yes. () () Plne (b) Upper hemisphere centered t the origin of rdius 3. (c) Rotte z = 1 bout z-xis. y (d) Cylindricl surfce, trnslte the prbol z = y in yz-plne long x-xis. (3) () x = (x, y, y x ) x x x y = (x, y, 1). Alterntively, consider F = z y + x nd find the grdient F to be (x, y, 1). At x = 4, y = 5 this vector is ( 8, 10, 1). So the tngent plne is 8(x + 4) 10(y 5) + z 9 = 0 8x 10y + z = 9 or 8x + 10y z = 9. (b) Consider F = x +y +3z 1 = 0 nd find F = (x, 4y, 6z). At (4, 1, 1), F = (8, 4, 6). The eqution of the plne is 8(x 4) 4(y+1)+6(z 1) = 0 8x 4y+6z = 4 4x y+3z = 1. (c) x = (u + v, 3u, u v) x u x v = ( 6u,, 6u). Then note tht t (, 3, 0), the vlues of prmeters re u = 1 nd v = 1 so the norml vector is (-6,,-6) The eqution of the plne is 6(x ) + (y 3) 6(z 0) = 0 6x + y 6z = 6 3x y + 3z = 3. (d) You cn prmetrize the cylinder s x = ( cos t, y, sin t) nd clculte x t x y to be ( cos t, 0, sin t). Note tht t (0, 3, ), the vlues of prmeters re t = π nd y = 3 so the norml vector is (0, 0, ). The eqution of the plne is 0(x 0) + 0(y 3) (z ) = 0 z =. Curvture nd Theorem Egregium For the concept of the curvture of curve γ to be defined, we hd to ensure tht we cn define unit-length tngent vector t every point. This condition ws ensured by requiring tht the derivtive dγ 0. Anlogously, for surfces we wnt to insure tht the tngent plne t every point is defined dt (i.e. tht is not collpsed into line or point). Since the norml vector to the tngent plne of prmetric surfce x is given by x x, we wnt to impose condition tht gurntees tht this u v vector is non-zero i.e., x u x v 0. Coordinte Ptches. The condition x x x x 0 gurntees tht the vectors nd re not u v u v on the sme line. Thus, they re linerly independent nd they constitute bsis of the tngent plne nd every other vector in the tngent plne cn be represented s liner combintion of these two vectors. 1 1 Liner Algebr bckground. Consider two vectors v 1 nd v tht do not lie on the sme line. In this cse, we sy tht v 1 nd v re linerly independent. Consider lso the plne determined by these two vectors. For rbitrry vector v in the plne, we cn consider the projection of v in direction of v 1. This projection is multiple of v 1. Let denote the multipliction fctor. Consider lso the projection of v in direction of v nd let b denote the the multipliction fctor. Thus, v = v 1 + bv The sum v 1 + bv is clled liner combintion of v 1 nd v. This shows tht every vector in the plne tht we 5

6 Thus, we shll consider just surfces such tht round every point prmetric equtions x = x(u, v), y = y(u, v), z = z(u, v) with the following properties cn be found. The functions x = x(u, v), y = y(u, v), z = z(u, v) re continuous in both vribles (thus, there re no gps or holes), one-to-one with continuous inverses (this lst condition gurntees properness ). The prtil derivtives of x = x(u, v), y = y(u, v), z = z(u, v) re continuous (thus, there re no corners or shrp turns). The cross product x x is not equl to 0 (thus, the tngent plne t ech point is not u v collpsed into line or point). We refer to those surfces s proper coordinte ptches. Note tht it my not be possible to describe the whole surfce with single coordinte ptch but it will lwys be possible to cover the entire surfce by ptching severl different coordinte ptches together. So, you cn think of coordinte ptches s bsic surfces tht crete rbitrry surfce. At the moment, we will not go into the forml definition of ptching but we will return to it lter. We present the informl ide of curvture t point P 0 on the surfce. We shll mke this ide more precise during the course of the semester. 1. Tke n rbitrry vector v of unit length in the tngent plne t P. () Consider the plne determined by v nd the norml vector of the tngent plne. This plne is perpendiculr to the tngent plne nd intersects the surfce in curve γ. The curve γ is clled the norml section t P in the direction of v. consider cn be expressed s liner combintion of v 1 nd v. In this cse, we sy tht v 1 nd v generte the plne. Linerly independent vectors tht generte plne re clled bsis. Considertion of projections bove demonstrtes tht ny two linerly independent vectors in plne constitute bsis of the plne. For exmple, vectors (1, 0) nd (0, 1) re bsis of xy-plne (spce R ): these two vectors re not coliner nd every vector (x, y) is the liner combintion x(1, 0) + y(0, 1). The sme concepts cn be defined in three-dimensionl spce. ny three vectors v 1, v nd v 3 tht do not lie on the sme plne re sid to be linerly independent. Any other vector v cn be expressed s sum of its projections in directions of v 1, v nd v 3 v = v 1 + bv + cv 3 i.e. s liner combintion of v 1, v nd v 3. Thus, v 1, v nd v 3 generte the spce. Linerly independent vectors tht generte the spce re clled bsis. By considering projections, ny three linerly independent vectors in spce re bsis of the spce. For exmple, vectors (1, 0, 0), (0, 1, 0) nd (0, 0, 1) re bsis of R 3 : they re not in the sme plne, nd every vector (x, y, z) is the liner combintion x(1, 0, 0) + y(0, 1, 0) + z(0, 0, 1). Another exmple of bsis of R 3 re the vectors T, N nd B of the moving frme of curve t ny of its points. 6

7 (3) Compute the curvture κ of γ t P. The norml curvture in the direction of v denoted by κ n (v) is tken to be κ n (v) = ±κ. If the norml vector of the curve nd the norml vector of the tngent plne re on the sme side of the tngent plne, then κ n (v) = κ. If they re on opposite sides of the tngent plne, then κ n (v) = κ. Otherwise κ n (v) = 0. Exmples. (1) Norml curvture of plne in ny direction is 0. This is becuse ny plne perpendiculr to the given plne intersects it in stright line so ll norml sections re lines nd, thus, hve zero curvture. () Absolute vlue of the norml curvture of sphere with rdius is 1. This is becuse ny plne perpendiculr to the sphere t point on it intersects the sphere in gret circle through the point. So, ll norml sections re circles of rdii with the curvture of 1. Thus, the norml curvture is ± 1. (3) Consider the cylinder x + y =. At ny point, plne perpendiculr to the tngent plne nd not prllel to the centrl xis intersects the cylinder in n ellipse. The curvture of n ellipse in the norml section chnges when v chnges so the norml curvture κ n is not constnt lso. There re two choices of v which stnd out: Let v 1 be vector in tngent plne perpendiculr to the centrl xis. The norml section in direction of v 1 is circle of rdius nd its curvture is 1. Let v be vector in the tngent plne prllel to the centrl xis. The norml section in direction of v is stright line nd its curvture is zero. These two directions determine the direction of lrgest nd smllest curving. Indeed, consider n rbitrry vector v in the tngent plne. Let θ be the ngle between v 1 nd v. Considering the right tringle in the figure on the right, we obtin tht the semi-xes of the ellipse in the norml section re nd. The cos θ curvture t the relevnt point is = cos θ. ( cos θ ) 7

8 Since cos θ is tking vlues between 0 nd 1, we obtin tht 0 κ n (v) = cos θ This shows tht κ n (v 1 ) = 1 is the mximl vlue of κ n(v) nd κ n (v ) = 0 is the miniml vlue. Principl curvtures. The bove exmple with the cylinder is interesting becuse it turns to be more generl thn one would imgine. Nmely, for every point on ny surfce, one cn choose orthogonl unit vectors v 1 nd v clled principl directions nd the norml curvtures determined by them will correspond to the mximl nd miniml vlues of the norml curvture κ n (v). These two vlues re denoted by κ 1 nd κ nd re clled principl curvtures. The product of the principl curvtures is clled the Gussin curvture K = κ 1 κ. 1. In the exmple with the cylinder, κ 1 = ± 1 nd κ = 0 so tht K = 0. The fct tht t every point of the cylinder there is direction in the tngent plne with the stright line s the norml section cuses the reltion K = 0. Note lso tht the cylinder cn be slit nd unrolled into flt sheet of pper without stretching or tering nd without ffecting the length ny curve. A surfce with this property will lso hve K = 0. This mens tht the geometry of the cylinder loclly is indistinguishble from the geometry of plne. In cses like this, we sy tht the two surfces re loclly isometric. Note tht globlly cylinder is very different from the plne, though. Following similr resoning, we cn deduce tht cone hs Gussin curvture 0. A sphere of rdius hs both the principl curvtures equl to 1, so the Gussin curvture is 1. Let us consider now the surfce given by z = y x clled hyperbolic prboloid. This surfce hs sddle point t the origin, nd the principl directions re in direction of x nd y xis. The norml sections in these directions re two prbols with norml vectors N 1 nd N hving the opposite directions. Thus, if the norml vector of the tngent plne hs the sme direction s N 1 then it hs the opposite direction to N nd vice vers. So, one principl curvture is positive nd the other is negtive resulting in negtive Gussin curvture t the origin. Hyperbolic prboloid hs K < 0 It cn be shown tht K is negtive throughout this curve. At this point, we note significnt difference between the curvture of curve nd the norml (nd Gussin) curvture of surfce: 8

9 while the curvture of curve is defined to hve just positive vlues, principl nd Gussin curvtures of surfces cn hve negtive vlues. The clcultion of curvture involves the surfce to be embedded in the spce so tht norml vectors nd norml sections could be considered. Thus, for locls on the surfce, these considertions would be incomprehensible. To understnd this rgument, consider the fct tht for people on the surfce of erth, the erth ppers to be flt, or the fct tht in our three-dimensionl world, we hve hrd time comprehending the four-dimensionl plne perpendiculr to our three-dimensionl spce (just s one-dimensionl curve hs two-dimensionl norml plne). A concept involving only mesurements on the surfce (conducted by locls ) is sid to be intrinsic while concept whose definition involves objects externl to the surfce is sid to be extrinsic. Thus, using the definition we presented, curvture is n extrinsic concept. If one is to generlize the concept of curvture to higher dimensions, in prticulr the curvture of our physicl spce, we would hve to be ble to describe curvture intrinsiclly. In prticulr, to determine the curvture of our physicl spce, we do not wnt to rely on fourdimensionl spce. Also, to define the curvture of Einstein four-dimensionl spce-time universe, we do not wnt to rely on five-dimensionl spce. Locls view surfce to be flt Fortuntely for locls, the crowning chievement of theory of surfces sttes tht the curvture cn be clculted intrinsiclly. This mens tht the Gussin curvture of surfce cn be determined entirely by mesuring ngles, distnces nd their rtes on the surfce itself, without further reference to the prticulr wy in which the surfce is embedded into the three-dimensionl spce. This result, proved by Crl Friedrich Guss, is considered to be one of foundtionl results in differentil geometry. It is usully referred to s Theorem Egregium (Ltin for remrkble or extrordinry theorem). In mthemticl lnguge, the theorem lso implies tht the Gussin curvture is invrint under locl isometry. This mens tht ny bending of surfce (without stretching or tering) does not impct the Gussin curvture. The principl curvtures do not shre the property of Gussin curvture given by Theorem Egregium the principl curvtures do vry with bending. The fct tht their product does not vry with bending mkes Theorem Egregium even more remrkble. We devote the reminder of our study of differentil geometry to ccomplishing the following three gols. Gol 1 Develop pprtus tht completely describes surfces. This will be nlogous to Serret-Frenet pprtus nd moving frme of curve nd will led us to first nd second fundmentl forms. Gol Understnd the sttement of Theorem Egregium in mthemticl terms. This will require considertion of geodesics nd the curvture tensor. Gol 3 Theorem Egregium llows the concept of curvture to be generlized to higher dimensions. Two-dimensionl surfces generlize to n-dimensionl mnifolds, defined for ny n nd the concept of the curvture of surfces generlizes to the curvture of mnifold. This will 9

10 enble you to understnd the lnguge used in specil nd generl reltivity. It will lso enble you to generlize the content of this course to higher dimensions. Theorem Egregium lso implies tht if two surfces hve different vlues of Gussin curvture, thn one cnnot be trnsformed into nother without tering or crumpling. To further motivte our study, we list severl corollries of this fct. A sphere (with K > 0) nd plne (with K = 0) cnnot be morphed one into nother. Thus, piece of pper cnnot be bent onto sphere without crumpling. As opposed to the cylinder (with K = 0), the sphere (with K > 0) cnnot be unfolded into flt surfce. Thus, if one were to step on n empty egg shell, its edges hve to split in expnsion before being flttened. An ornge peel cn be flttened just with tering or stretching. As consequence of previous observtions, the Erth cnnot be displyed on mp without distortion. Thus, no perfect mp of Erth cn be creted, even for portion of the Erth s surfce nd every crtogrphic projection necessrily distorts t lest some distnces. This fct is of enormous significnce for crtogrphy. Every distinct mp projection distorts in distinct wy. The study of mp projections is the chrcteriztion of these distortions. Merctor projection A frequently used projection, Merctor projection, preserves ngles but fils to preserve re (tht is why the res round north nd south pole look disproportiontely lrge compred to the res further wy from the poles). The controversy surrounding the Merctor projections rose from politicl implictions of mp design since representing some countries lrger thn the others my implied tht some re less significnt. Norml nd Trnsverse Merctor projections Another projection used in some cses is Gll- Peters projection (you cn see it in some world mps on irplnes). On this projection res of equl size on the globe re lso eqully sized on the mp. This hs consequence tht res round the equtor looks elongted when compred to res with lrger geogrphicl width. 10 Gll-Peters projection

11 Merctor nd Gll-Peters with their deformtions When trying to preserve precious toppings on slice of pizz, you re using Theorem Egregium too: you bend slice horizontlly long rdius so tht non-zero principl curvtures re creted long the bend, dictting tht the other principl curvture t these points must be zero. This cretes rigidity in the direction perpendiculr to the fold nd it prevents the toppings from flling off. Theorem Egregium lso implies tht we cn mesure the curvture of the Erth without leving the surfce (for exmple in n irplne to observe the curving) just mesuring the distnces nd ngles on the surfce of the Erth. Prctice Problems.. Determine 1. The men curvture is defined s the men of the principl curvtures H = κ1 +κ the bsolute vlue of the men curvture of the surfces discussed in this section: plne, sphere of rdius nd cylinder x + y =.. Find the Gussin curvture of the hyperbolic prboloid z = y x t the origin using tht the principl directions re the directions of positive x nd y xis. 3. Find the Gussin curvture of ellipsoid x + y + zc = 1 t the end points of the three semib xes (±, 0, 0), (0, ±b, 0) nd (0, 0, ±c). Ellipsoid 4. A qudrtic surfce is ny surfce given by eqution x + by + cz + dxy + exz + f yz + gx + hy + iz + j = 0. This clss includes the following surfces: ellipsoid ( x + yb + zc = 1), ellipticl prboloid ( x + yb = z), hyperbolic prboloid ( x yb = z), hyperboloid of one sheet ( x + yb zc = 1) nd hyperboloid of two sheets ( x + yb zc = 1). Ellipticl nd hyperbolic prboloids nd hyperboloids of one nd two sheets 11

12 By mking suitble chnge of vribles to eliminte some terms, ny qudrtic surfce cn be put into certin norml form. It turns out tht there re 16 such norml forms. Of these 16 forms, the bove five surfces re non-degenerte nd remining eleven re degenerte: cones ( x + y z = 0), cylindricl surfces (elliptic, hyperbolic nd prbolic cylinder), plnes, b c lines, points or even no points t ll. Using rgument similr to those used to show tht the Gussin curvture of cylinder is 0, deduce tht K of ll the degenerte qudrtic surfces is 0. Then determine the sign of Gussin curvture for five non-degenerte qudrtic surfces. 5. A torus is surfce obtined by revolving one circle long the other circle creting doughnutlike shpe. Consider revolving circle (x ) +z = b in xz-plne long the circle x +y = in xy-plne. Assume tht > b so tht the doughnut tht you obtin relly hs hole in the middle. Clculte the Gussin curvture t ny point on the outer circle (obtined by revolving the point ( + b, 0, 0) bout z-xis) nd t ny point on the inner circle (obtined by revolving the point ( b, 0, 0) bout z-xis. Torus Solutions. (1) H = 0+0 = 0 for ny plne, H = = 1 for the sphere of rdius, nd H = 0+ 1 = 1 for the cylinder x + y =. () In section on curves, we computed the curvture of prbol y = x t x = 0 to be. In similr mnner we cn obtin the curvture of prbol y = x to be s well. The norml sections of z = y x re two prbols in xz nd yz plnes. In xz plne, y = 0 nd so z = x nd in yz plne x = 0 nd z = y both with curvtures. The two norml vectors hve the opposite direction so the the two principl curvtures re nd -. Thus, κ 1 =, κ = nd K = 4. (3) Let us clculte the curvture of ellipse x + y = 1 t (, 0) nd (0, b) first. Here the curve b cn be prmetrized s γ = ( cos t, b sin t). Then γ = ( sin t, b cos t, 0), γ = ( cos t, b sin t, 0) γ γ = (0, 0, b). Thus γ = sin t + b cos t nd γ γ b = b nd so κ = ( sin t+b cos t) 3/. At (, 0) the vlue of prmeter t is 0 nd t (0, b) the vlue of prmeter t is π b. Thus κ(0) = = b 3 b nd κ( π) = b = b. 3 At (±, 0, 0), the norml sections re in xy nd xz plnes. In xy plne the norml section is the ellipse x + y = 1 with curvture t (±, 0). In xz plne the norml section is the ellipse b b x + z = 1 with curvture t (±, 0). The norml vectors hve the sme directions so the two c c principl curvtures hve the sme signs. Thus the Gussin is K =. b c At (0, ±b, 0), the norml sections re in xy nd yz plnes. In xy plne the norml section is the ellipse x + y = 1 with curvture b t (0, ±b). In yz plne the norml section is the ellipse b y + z = 1 with curvture b. b c c 1

13 The norml vectors hve the sme directions so the two principl curvtures hve the sme signs. Thus the Gussin is K = b. On similr c mnner, we obtin tht the Gussin curvture t (0, 0, ±c) is K = c. b (4) K > 0 for ellipsoid. For ellipticl prboloid of the form z = x + y K > 0. K < 0 b for hyperbolic prboloid nd for hyperboloid of one sheet, K > 0 for hyperboloid of two sheets. Hyperbolicl Prboloids (5) At point ( + b, 0, 0), the norml sections re in xy plne nd xz plne. In xy plne, the 1 norml section is the circle of rdius + b so its curvture is. In xz plne the norml section is +b the circle of rdius b so its curvture is 1. The norml vectors hve the sme direction. Hence, the b 1 Gussin curvture is positive nd equl to. b(+b) At point ( b, 0, 0), the norml sections re in xy plne nd xz plne s well. In xy plne, the 1 norml section is the circle of rdius b with curvture. In xz plne the norml section is b the circle of rdius b with curvture 1. The norml vectors hve the opposite direction. Hence, the b 1 Gussin curvture is negtive nd equl to. b( b) Using this exmple, we cn deduce tht on the outer prt of the torus (obtined by revolving the right hlf of circle (x ) + z = b bout z-xis) the Gussin curvture is positive nd on the inner prt of the torus (obtined by revolving the left hlf of circle (x ) + z = b bout z-xis) the Gussin curvture is negtive. This implies the not so obvious fct tht the Gussin curvture on the top nd bottom circles (obtined by revolving points (, 0, b) nd (, 0, b) bout z-xis) is zero. 13