MA304 Differential Geometry

Transcription

1 MA304 Differential Geoetry Hoework 4 solutions Spring 018 6% of the final ark 1. The paraeterised curve αt = t cosh t for t R is called the catenary. Find the curvature of αt. Solution. Fro hoework question 1 the curvature of α at t is given by kt = α α. We α 3 introduce a third coordinate and write αt = t cosh t 0. Then α = 1 sinh t 0 and α = 0 cosh t 0 so α α = 0 0 cosh t and α α = cosh t. Also α = 1 +sinh t = cosh t. Therefore kt = α α α 3 = cosh t cosh 3 t = 1 cosh t = sech t.. Let fx y z = x + y + z 1. a Locate the critical points and critical values of f. b For what values of c is the set fx y z = c a regular surface? c Answer the questions of parts a and b for the function fx y z = xyz. Solution. a A point p R 3 is a critical point of fx y z = x + y + z 1 if and only if f x = f y = f z = 0 at p. We have f x = f y = f z = x + y + z 1 and thus f x f y and f z vanish siultaneously if and only if x + y + z = 1. That is p is a critical point if and only if p lies on the plane x + y + z = 1. When x + y + z = 1 we have fx y z = 0 so 0 is the only critical value of f. b Clearly the range of values taken by f is [0 +. We have fx y z = 0 if and only if x+y+z = 1 which is a plane. Also if c > 0 then fx y z = c if and only if x+y+z = ± c+1 which consists of two parallel planes. Therefore the set fx y z = c is a regular surface if and only if c [0 +. c Siilar to part a we have f x = yz f y = xz and f z = xyz. We see that f x = f y = f z = 0 if and only if either z = 0 or z 0 and x = y = 0. In other words the critical points of f lie either on the xy-plane or on the z-axis. In either case we have fx y z = 0 and thus 0 is the only critical value. We see that the range of values taken by f is R for exaple we can fix y = z = 1 and let x vary over R. Thus every real nuber c R \ {0} is a regular value of f. By Proposition 3 in Chapter the surface given by fx y z = c that is xyz = c is a regular surface. On the other hand we have xyz = 0 if and only if x = 0 or y = 0 or z = 0. That is the set given by xyz = 0 is the union of the xy xz and yz-planes and this is clearly not a regular surface. Thus the set fx y z = xyz = c is a regular surface if and only if c R \ {0}. 3. Let H = {x y z R 3 : x y + z = 1} be the hyperboloid of two sheets. a Find a paraeterisation for H. b Find the coefficients of the first fundaental for with respect to the paraeterisation that you found in part a. c Find the area of H when restricted to 0 z <. 1

2 Solution. a We have x y + z = 1 z = ± 1 + x + y. Since 1 + x + y > 0 and 1 + x + y < 0 we see that H consists of two disjoint surfaces and they can be paraeterised by where u = x v = y R. xu v = u v 1 + u + v and xu v = u v 1 + u + v b We find the coefficients of the first fundaental for for both xu v and xu v. x u = 1 0 u1 + u + v 1/ x v = 0 1 v1 + u + v 1/ E = x u x u = 1 + u 1 + u + v 1 F = x u x v = uv1 + u + v 1 G = x v x v = 1 + v 1 + u + v 1. x u = 1 0 u1 + u + v 1/ x v = 0 1 v1 + u + v 1/ Ẽ = x u x u = 1 + u 1 + u + v 1 F = x u x v = uv1 + u + v 1 G = x v x v = 1 + v 1 + u + v 1. c When we restrict to 0 z < we are clearly considering the upper sheet of the hyperboloid paraeterised by xu v. Note that the planes z = c for 1 c intersect the hyperboloid in circles of increasing radius with z = corresponding to a circle of radius 1. We see that to find the required area the region of integration R in the uv-plane is given by u + v < 1. The area is A = R EG F du dv = R 1 + u 1 + u + v + v 1 + u du dv. + v We convert to polar coordinates r θ noting that we ust ultiply by the Jacobian of r in the new integrand. We have π 1 1 A = 1 + r r dr dθ = π r 1 + r r dr = π 1 t dt 0 0 where t = 1 + r. It is possible to copute the last integral above but it is rather coplicated. Fro a nuerical approxiation for the integral is Thus the required area is approxiately 1.141π = Consider the paraboloid z = x + ay where a > 0. Show that at p = the unit vectors and are eigenvectors of dn p with eigenvalues and a respectively assuing that N is pointing outwards fro the region bounded by the paraboloid. Solution. We paraeterise the paraboloid as We have xu v = u v u + av u = x u = v R. x u = 1 0 u 0 x v = 0 1 av Since the unit noral Nu v is pointing outwards fro the region bounded by the paraboloid we have Nu v = x v x u x v x u. Since x v x u = u av 1 and x v x u = 4u + 4a v + 1 we can write the unit noral as Nx y z = x ay 1 4x + 4a y

3 Now the curve αt = t 0 t lies on the paraboloid. We have α0 = p and α t = 1 0 t so that α 0 = Thus t βt = N αt = 4t + 1 1/ 0 1 4t + 1 1/ 4t β + 1 1/ 8t 4t + 1 1/ t = 4t 0 4t4t + 1 3/ + 1 dn p = β 0 = 0 0. Therefore is an eigenvector of dn p with eigenvalue. Siilarly the curve αt = 0 t at lies on the paraboloid. We have α0 = p and α t = 0 1 at so that α 0 = Thus at βt = N αt = 0 4a t + 1 1/ 1 4a t + 1 1/ β t = 0 a4a t + 1 1/ 8a 3 t 4a t + 1 1/ 4a t 4a t4a t + 1 3/ + 1 dn p = β 0 = 0 a 0. Therefore is an eigenvector of dn p with eigenvalue a. 5. Show that if a regular orientable surface is tangent to a plane along a curve then the points of this curve are either parabolic or planar. Solution. Let S be the surface. Let p be a point on the curve and suppose that the curve is paraeterised near p as α : ε ε S with α0 = p. The given condition on the curve α eans that all points on the trace α ε ε S lie in T p S and they have a constant unit noral N to S. Thus the curve β = N α is a constant point in the unit sphere S over ε ε. Moreover the vector w = α 0 lies on T p S and thus dn p w = β 0 = 0. Therefore w is an eigenvector of dn p with eigenvalue 0 and detdn p = 0. We have p is a parabolic or a planar point and the assertion follows since p is an arbitrary point on the given curve. 6. Describe the region of the unit sphere covered by the iage of the Gauss ap of the following surfaces. a Paraboloid of revolution z = x + y. b Hyperboloid of revolution x + y z = 1. c Catenoid x + y = cosh z. Solution. In all three parts let x y z and X Y Z denote the coordinates of the surface in consideration and of the unit sphere S respectively. We have X + Y + Z = 1. a Let Nx y z be the unit noral to the paraboloid z = x +y at the point x y z pointing into the region bounded by the paraboloid. Fro question 4 with a = 1 we have Nx y z = x y 1. 4z + 1 4z + 1 4z + 1 Since z = x + y the range of z is [0 +. Thus the range of Z = 1 4z+1 is 0 1] and the noral Nx y z ust necessarily lie in the upper heisphere of S where 0 < Z 1. On the 3

4 other hand we show that any point X Y Z S with 0 < Z 1 can be attained by soe noral Nx y z with z = x + y. Given such X Y Z we choose z so that Z = 1 4z+1 in other words z = 1 Z. We then choose x and y such that X = x 4Z 4z+1 and Y = other words x = X Z and y = Y Z. Note that we have x + y = X +Y the point x y z lies on the paraboloid z = x + y. 4Z = 1 Z 4Z The region of the sphere S is therefore the upper heisphere where 0 < Z 1. y 4z+1 in = z so that b Let Nx y z be the unit noral to the hyperboloid x + y z = 1 at the point x y z pointing away fro the region bounded by the hyperboloid. We will use the ethod in hoework 3 question to copute Nx y z. The hyperboloid is a surface of revolution which can be obtained by rotating the curve x = + z + 1 > 0 in the xz-plane about the z-axis. Let p = x 0 y 0 z 0 be a point on the hyperboloid and let 0 u 0 < π be the longitude of p that is u 0 is the angle that the vector x 0 y 0 in the xy-plane akes with the positive x-axis. The parallel through p is the circle x + y = z0 + 1 and z = z 0 so a tangent vector to the parallel at p is y 0 x 0 0. Next a paraeterisation αv for the eridian through p is αv = v + 1 1/ cos u 0 v + 1 1/ sin u 0 v v = z R. We have α v = v + 1 1/ v cos u 0 v + 1 1/ v sin u 0 1. Since v = z 0 at p the tangent vector to the eridian at p is α z 0 = z / z 0 cos u 0 z / z 0 sin u 0 1. Since x 0 = z0 + 11/ cos u 0 and y 0 = z0 + 11/ sin u 0 we have y 0 x 0 0 α z 0 = x 0 y 0 y 0 z / z 0 sin u 0 x 0 z / z 0 cos u 0 = x 0 y 0 x 0 + y 0 z 0 z0 + 1 = x 0 y 0 z 0 x so that the unit noral to the hyperboloid at p is 0 y 0 z 0 = x 0y 0 z 0. Therefore for a x 0 +y 0 +z 0 1/ z0 +11/ point x y z of the hyperboloid the unit noral is x Nx y z = z + 1 y z + 1 z. z + 1 Since z = ± x + y 1 the range of z is R. Since the function z z +1 is continuous and strictly decreasing and z ± 1 z +1 as z it follows that the range of Z = z z +1 is 1 1 and the noral Nx y z ust necessarily lie in the region of S where 1 < Z < 1. On the other hand we show that any point X Y Z S with 1 < Z < 1 can be attained by soe noral Nx y z with x + y z = 1. Given such X Y Z we choose z so that Z = z. Note that z is uniquely deterined fro Z since z is a bijection z +1 fro R to 1 1. We then choose x and y such that X = z +1 x z +1 and Y = y z in other +1 words x = X z + 1 and y = Y z + 1. Note that we have x + y = X + Y z + 1 = 1 Z z + 1 = z + 1 Z z + 1 = z + 1 so that the point x y z lies on the hyperboloid x + y z = 1. The region of the sphere S is therefore the region where 1 < Z < 1. In other words it is the region of the sphere S where the latitude is in π 4 π 4. c Let Nx y z be the unit noral to the catenoid x + y = cosh z at the point x y z pointing away fro the region bounded by the catenoid. We will use a siilar ethod as in part b to copute Nx y z. The catenoid is a surface of revolution which can be obtained 4

5 by rotating the curve x = cosh z > 0 in the xz-plane about the z-axis. Let p = x 0 y 0 z 0 be a point on the catenoid and let 0 u 0 < π be the longitude of p as before. The parallel through p is the circle x + y = cosh z 0 and z = z 0 so a tangent vector to the parallel is y 0 x 0 0. Next a paraeterisation αv for the eridian through p is αv = cosh v cos u 0 cosh v sin u 0 v v = z R. We have α v = sinh v cos u 0 sinh v sin u 0 1. Since v = z 0 at p the tangent vector to the eridian at p is α z 0 = sinh z 0 cos u 0 sinh z 0 sin u 0 1. Now y 0 x 0 0 α z 0 = x 0 y 0 y 0 sinh z 0 sin u 0 x 0 sinh z 0 cos u 0. Since x 0 = cosh z 0 cos u 0 and y 0 = cosh z 0 sin u 0 the third coordinate above siplifies as sinh z 0 x 0 cos u 0 + y 0 sin u 0 = sinh z 0x 0 + y 0 cosh z 0 = sinh z 0 cosh z 0 x so that the unit noral to the catenoid at p is 0 y 0 sinh z 0 cosh z 0 = x 0y 0 sinh z 0 cosh z 0 x 0 +y 0 +sinh z 0 cosh z 0 1/ cosh z 0 by using x 0 + y 0 = cosh z 0. For a point x y z of the catenoid the unit noral is x Nx y z = cosh z y cosh z tanh z. Since cosh z = x + y the range of z is R. Since the function tanh z is continuous and strictly decreasing with range 1 1 it follows that the range of Z = tanh z is 1 1 and the noral Nx y z necessarily cannot be or On the other hand we show that any point X Y Z S \ { } can be attained by soe noral Nx y z with x + y = cosh z. Given such X Y Z we choose z so that Z = tanh z. Note that z is uniquely deterined fro Z since tanh z is a bijection fro R to 1 1. We then choose x and y such that X = x cosh z and Y = y cosh z in other words x = X cosh z and y = Y cosh z. Note that we have x + y = X + Y cosh 4 z = 1 Z cosh 4 z = 1 tanh z cosh 4 z = sech z cosh 4 z = cosh z so that the point x y z lies on the catenoid x + y = cosh z. The region of the sphere S is therefore S \ { }. Finally we reark that in all three parts if we chose the unit noral Nx y z to point in the opposite direction then we would obtain the regions of the sphere S which are the iages of the ones that we have obtained under the antipodal ap on S. Naely we obtain the lower heisphere where 1 Z < 0 for part a and the sae regions in parts b and c. 7. Optional for bonus Let S be a regular orientable surface. Let λ 1... λ be the noral curvatures at p S along directions aking angles 0 π/... 1π/ with a principal direction. Prove that λ λ = H where H is the ean curvature at p. Solution. The assertion does not hold for = 1. We prove it for 3. We recall that if the surface S is given a suitable orientation and S has principal curvatures k 1 k at a point p then there is an orthonoral basis {f 1 f } in the tangent plane T p S such that k 1 and k are attained in the directions of f 1 and f. Moreover Euler s forula states that if v is a unit vector in T p S and θ is the angle fro f 1 to v then the noral curvature k n of S in the direction of v is given by k n = k 1 cos θ + k sin θ. 5

6 Suppose that λ 1... λ as given in the question ake angles of 0 π/... 1π/ with the direction of k 1. Then λ λ = k 1 cos 0 + cos π + + cos 1π + k sin 0 + sin π + + sin 1π. We show that both of the series on the right are equal to /. Then the right hand side will siplfy to H. We have cos 0 + cos π + + cos 1π 4π 14π = + cos 0 + cos + + cos sin 0 + sin π + + sin 1π 4π 14π = cos 0 cos cos and so it suffices to show that 4π 14π cos 0 + cos + + cos = 0. 1 We note that this su is the real part of the coplex nuber z = e 0 + e 4πi/ + e 8πi/ + + e 14πi/ where for the ters in this su we start with e 0 and then each successive ter is obtained by rotating the previous ter by the angle 4π/. Now observe that if we have a set of at least two coplex nubers with unit odulus distributed evenly around the unit circle then their su is 0 since we can rearrange the coplex nubers to for the sides of a regular polygon in the case of two coplex nubers we have a degenerate -gon. If 4 is even then we have e 0 = e /4πi/ e 4πi/ = e /+14πi/... etc. so that z = e 0 + e 4πi/ + e 8πi/ + + e / 14πi/. The ters in the su are coplex nubers distributed evenly around the unit circle and hence z = 0. If 3 is odd then we have e πi/ = e +1/4πi/ e 6πi/ = e +1/+14πi/... etc. so that z = e 0 + e πi/ + e 4πi/ + + e 1πi/. Again the ters in the su are coplex nubers distributed even around the unit circle and hence z = 0. Therefore the identity 1 holds and we have λ λ = H as required. Finally suppose that λ 1... λ ake angles of 0 π/... 1π/ with the direction of k. Then λ 1... λ ake angles of π/ π/ + π/... 1π/ + π/ with the direction of k 1 and thus as before λ λ = k 1 cos π + π cos + π + + cos 1π + k sin π + sin π + π + + sin 1π + π = k 1 sin 0 + sin π + + sin 1π + k cos 0 + cos π + + cos 1π = k 1 + k = H. 6 + π

7 8. Optional for bonus Let 4 be an even integer. Give an explicit construction of a differentiable siple closed curve in the plane with exactly vertices. [Hint: The closed curve does not need to be convex.] Solution. We assue throughout that 6 is even since for the case = 4 we ay consider an ellipse see hoework question b. We consider an epitrochoid as our choice for the closed curve α. The trace of α is obtained as follows. Fix a circle C with radius R and centre O = 0 0. Initially place a circle Γ with radius r and centre A = R + r 0 and fix a point P in Γ at distance d 0 fro A towards O. We then roll Γ around the outside of C and obtain the curve traced out by P. Figure 1 below shows the case when R = 3 r = 1 and d = 1/. Figure 1 The paraeterisation for α by introducing a third coordinate is R + r R + r αt = R + r cos t d cos t R + r sin t d sin t 0 r r where t [0 π]. We reark that t is not the polar angle. Let n = / 3. We set R = n r = 1 and we will choose d later. Then αt = n + 1 cos t d cosn + 1t n + 1 sin t d sinn + 1t 0 α t = n + 1 sin t + dn + 1 sinn + 1t n + 1 cos t dn + 1 cosn + 1t 0 α t = n + 1 cos t + dn + 1 cosn + 1t n + 1 sin t + dn + 1 sinn + 1t 0. Recall that the curvature of α is given by kt = α α α 3. We have α α = [ n + 1 sin t + dn + 1 sinn + 1t ][ n + 1 sin t + dn + 1 sinn + 1t ] [ n + 1 cos t dn + 1 cosn + 1t ][ n + 1 cos t + dn + 1 cosn + 1t ] = n + 1 [ sin t + d sinn + 1t ][ sin t + dn + 1 sinn + 1t ] [ cos t d cosn + 1t ][ cos t + dn + 1 cosn + 1t ] = n d n + 1 dn dsin t sinn + 1t + cos t cosn + 1t = n d n + 1 dn + cos nt 7

8 and α = n + 1 sin t + dn + 1 sinn + 1t + n + 1 cos t dn + 1 cosn + 1t = n + 1 [ sin t + d sinn + 1t + cos t d cosn + 1t ] = n + 1 [ 1 + d dsin t sinn + 1t + cos t cosn + 1t ] = n d d cos nt. For α α to be differentiable we choose d so that the ter in is non-negative. It suffices that 1 + d n + 1 dn + 0 or equivalently d 1n + 1d 1 0. We choose d = 1 n+1 so that 1 + d n + 1 dn + = 0. Then α α = n + 1 dn + dn + cos nt = n + 1n + 1 cos nt α = n n + 1 cos nt > 0 kt = n + 1n + 1 cos nt n n + 1 cos nt 3/. By using the quotient rule to differentiate kt we see that k t = 0 the resulting nuerator is equal zero. Writing for n n + 1 cos nt > 0 we obtain the nuerator as [ n + 1n + n sin nt 3/ 1 cos nt 3 ] 1/ nn + 1 sin nt = nn + 1n + 1/ sin nt [ 3n + 11 cos nt ]. Thus k t = 0 when either sin nt = 0 or 3n + 11 cos nt = 0. For the latter case we see that 3n + 11 cos nt = n n + 1 cos nt 3n + 11 cos nt = n + 1n n + 1 cos nt > 0 for n 3 so that this case cannot occur. Therefore the vertices occur precisely when sin nt = 0 that is t = 0 π/n π/n... n 1π/n and so α has precisley n = vertices. Figure below shows the cases when = where the circle C and the vertices are indicated. We see that the curvature kt alternates between iniu and axiu at the vertices. Figure Soe online resources can be found at http: //athworld.wolfra.co/epitrochoid.htl and u0bupudip5. 8