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Solutions

Chapter 1

1.1 It is a parametrisation of the part of the parabola with x ~ O.1.2 (i) -y(t) = (sect,tant) with -7r/2 < t < 7r/2 and 7r/2 < t < 37r/2.

(ii) -y(t) = (2cost ,3sint).1.3 (i) x 1- y =1.

(ii) y = (lnx)2 .1.4 (i) i'(t) = sin2t(-I, 1).

(ii) i'(t) = (e", 2t).1.5 i'(t) = 3 sin t cos t( - cos t, sin t) vanishes where sin t = 0 or cos t = 0, i.e.

t = n7r /2 where n is any integer. These points correspond to the fourcusps of the astroid.

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282 Elementary Differential Geometry

1.6 (i) Let OP make an angle 0 with the positive z-axis. Then R has coordinates 'Y(O) = (2acotO,a(1 - cos20)).(ii) From x = 2acotO, y = a(l- cos20), we get sin2

(} = y/2a, cos2 (} =cot 2

() sin2() = x 2 y/8a 3 , so the cartesian equation is y/2a + x 2 y/8a 3 = 1.

1.7 When the circle has rotated through an angle t, its cent re has moved to(at , a), so the point on the circle initially at the origin is now at the point(a(t - sin t), a(l - cos t)) .

ot

1.8 Let the fixed circle have radius a, and the moving circle radius b (so thatb < a in the case of the hypocycloid), and let the point P of the movingcircle be initially in contact with the fixed circle at (a, 0). When the movingcircle has rotated through an angle tp, the line joining the origin to thepoint of contact of the circles makes an angle 0 with the positive z-axis,where a(} = bip, The point P is then at the point

'Y(O) = «a + b)cosO - bcos(O + cp), (a + b)sin(} - bsin(O + tp))

=«a + b)cosO - bcos«a + b)O/b), (a + b)sinO - bsin«a + b)O/b))

in the case of the epicycloid,

and'Y(O) = «a - b)cosO + bcos(tp - 0), (a - b)sinO - bsin(tp - 0))

=«a - b)cosO + bcos«(a - b)O/b), (a - b)sinO - bsin((a - b)O/b))

in the case of the hypocycloid .

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1.9 A point (x, y, z) lies on the cylinder if x2 + y2 = 1/4 and on the sphereif (x + ~)2 + y2 + z2 = 1. From the second equation, -1 ~ z ~ 1 solet z = sin t . Subtracting the two equations gives x + t + sin2 t = i , sox = ~ - sin2 t = cos2 t - ~ . From either equation we then get y = sin t cos t(or y = - sin t cos t , but the two solutions are interchanged by t f--t 7r - t).

1.10 -y(t) = (e'{cos r - sint),et(sint + cos ej) so the angle (J between 'Y(t) and-y(t) is given by

(J 'Y.-Y e2t(cos2t-sintcost+sin2t+sintcost) 1cos = 11 'Y 1111 -y 11 = e2t«cos t - sin t)2 + (sin t + cos t)2) = 2'

and so (J = 7r/3.1.11 -y(t) = (1, sinh t) so 11 -y 11 = cosh t and the arc-length is s = f~ cosh u du =

sinh t.1.12 (i) 11 -y 11 2 = t(l + t) + t(l - t) + ~ = 1.

(ii) 11 -y 11 2 = ~: sin2 t + cos2 t + i5 sin2 t = cos2 t + sin2 t = 1.1.13 The cycloid is parametrised by 'Y(t) = a(t - sin t, 1 - cos t), where t is

the angle through which the circle has rotated. So -y = a(1 - cos t, sin t) ,11 -y 11 2 = a2(2 - 2 cos t) = 4a2 sin2 ~, and the arc-length is

1271" t t It = 271"2a sin 2dt = -4a cos 2 = 8a.

o t=o

1.14 (i) -y =sin2t(-I, 1) vanishes when t is an integer multiple of 7r/2, so 'Y isnot regular.(ii) Now 'Y is regular since -y ::j:. 0 for 0 < t < n /2.(iii) -y = (1, sinh t) is obviously never zero, so 'Y is regular.

1.15 x = r cos (J = sin2 (J, y = r sin (J = sin2 (Jtan (J , so the parametrisation interms of (J is (J f--t (sin2 (J , sin2 (J tan (J). Since (J f--t sin (J is a bijective smoothmap (-7r/2,7r/2) --7 (-1 ,1), with smooth inverse t f--t sin"! t , t = sin f isa reparametrisation map. Since sin2 (J = t2 , sin2 (J tan (J = t3/~, thereparametrised curve is as stated.

1.16 We have s = ft: 11 d'Y/du 11 du, S = fL 11 d;y/du 11 du. By the chain rule,

d'Y/du = (d;Y/du)(du/du), so s = ± ft: 11 d;y/du 11 (du/du) du = ±s, thesign being that of du/du .

1.17 If'Y(t) = (x(t),y(t),z(t» is a curve in the surface f(x ,y,z) =0, differentiating f(x(t), y(t), z(t» = 0 with respect to t gives xfx + iJfy + ifz = 0,so -y is perpendicular to (Ix, fy , fz). Since this holds for every curve in thesurface , (Ix , fy, fz) is perpendicular to the surface. The surfaces f = 0 and9 = 0 should intersect in a curve if the vectors (Ix, fy, fz) and (gx,gy,gz)are not parallel.

1.18 Let 'Y(t) = (u(t), v(t), w(t» be a regular curve in R 3. At least one of ü, V, tU

is non-zero at each value of t , Suppose that ü(to) ::j:. 0 and xo = u(to). As


in the 'proof' of Theorem 1.2, there is a smooth function h(x) definedfor x near Xo such that t = h(x) is the unique solution of x = u(t) foreach t near to. Then, for t near to, -y(t) is contained in the level curve!(x,y,z) = 9(X,y,Z) = 0, where !(x,y,z) = y - v(h(x» and 9(X,y,z) =z - w(h(x». The functions ! and 9 satisfy the conditions in the previousexercise, since (fx , !y, !z) = (-iJh', 1,0), (9x, 9y, 9z) = (-wh', 0,1), a dashdenoting d]dx.

1.19 Define 8(t) = tan rr{}(t)/2, where {} is the function defined in Exercise8.20. Then 8 is smooth, 8(t) = 0 if t ~ 0, and 8 : (0, (0) --t (0, (0) is abijection. The curve

{(8(t) ,8(t»

-y(t) = (-8(-t), 8(-t»

is a smooth parametrisation of y = [z],

if t ~ 0,if t ~ 0,

There is no regular parametrisation of y = [z]. For if there were, therewould be a unit-speed parametrisation ;y(t), say, and we can assume that;y(0) = (0,0) . The unit tangent vector-j would haveto be either ~(1, 1) or

- ~(1, 1) when x > 0, so by continuity we would have ly(O) = ±~(1, 1).

But, by considering the part x < 0 in the same way, we see that ly(O) =±~(1 , -1). These statements are contradictory.

Chapter 2

2.1 (i) -y is unit-speed (Exercise 1.12(i» so

It = 11 ;Y 11 = 11 (-41

(1 + t)-1/2, -41

(1 - t)-1/2, 0) 11 = 1V8(1 - t2 )

(ii) -y is unit-speed (Exercise 1.12(ii» so

1t=1/;Y1/=1I (-~cost,sint,~cost) 11= 1.

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(iii) Using Proposition 2.1,

_ 11 (1, sinh t, 0) x (0, cosh t, 0) 11 _ cosh t _ h2", - - -- - sec t.11 (1, sinh t, 0) 11 3 cosh" t

(iv) (-3 cos? t sin t, 3 sin2 t cost, 0) x (-3 cos3 t+6 costsin2 t, 6 sin t cos2 t3 sin3 t, 0) = (0,0, -9 sin2 t cos2 t), so

11 (0,0, -9 sin2 t cos2 t) 11 1", = 11 (-3 cos2 t sin t, 3 sin2 t cos t, 0) 11 3 = 31 sin t cos tl'

This becomes infinite when t is an integer multiple of 7l"/2, i.e, at the fourcusps (±I,O) and (0, ±I) of the astroid.

2.2 The proof of Proposition 1.4 shows that, if v(t) is a smooth (vector)function of t, then 11 v(t) 11 is a smooth (scalar) function of t provided v(t)is never zero. The result now follows from the formula in Proposition 2.1.The curvature of the regular curve 'Y(t) = (t, t3) is ",(t) = 6Itl/(I + 9t4)3/2,which is not differentiable at t = O.

2.3 Differentiate t .n, = 0 and use i = ",sns.2.4 If 'Y is smooth, t ="I is smooth and hence so is n, (since it is obtained by

applying a rotation to t) and i. So "'s = t .n, is smooth.2.5 The arc-length s = J 11 "I 11 dt = J 11 ekt (k cos t - sin t, k sin t + cos t) 11 dt

= JVk2 + Iekt dt = ~ekt + c, where cis a constant. Taking c = 0makes s ~ 0 as t ~ =foo if ±k > O.Since t = "1/11 "I 11= v'k;+l(kcost-sint,ksint+cost), we have

n, = v'k!+l (-ksint - cost,kcost - sint). So dt/ds = (dt/dt)/(ds/dt) =-Alt -let

:2+1 (-ksint-cost,kcost-sint) = Jk2+lnS so x, = I/ks . By Theorem2.1, any other curve with the same signed curvature is obtained from thelogarithmic spiral by applying a rigid motion .

2.6 (i) Differentiating 'Y = rt gives t = rt + "'srns. Since t and n, are perpendicular unit vectors, it follows that "'s = 0 and 'Y is part of a straightline.(ii) Differentiating ry = rns gives t = rn, + rns = rns - "'srt (Exercise2.3). Hence, r =0, so r is constant, and x, = -I/r, hence x, is constant.So 'Y is part of a circle.(iii) Write'Y = r(t cosO+ns sin 0). Differentiating and equating coefficientsof t and n, gives r cos0 - ",sr sin 0 =1, r sin 0 + ",sr cos0 =0, from whichr = cosO and ",sr = -sinO. From the first equation, r = scosO (we canassurne the arbitrary constant is zero by adding a suitable constant tos) so x, = -1/scotO. By Exercise 2.5, 'Y is obtained by applying a rigidmotion to the logarithmic spiral defined there with k = - cot O.

2.7 We can assurne that 'Y is unit-speed. Then "I>" = (1 - A"'s)t, and this isnon-zero since 1 - A",s > O. The unit tangent vector of 'Y>" is t, and the


arc-length s of '1>' satisfies ds/dt =1- )."'s. Hence, the curvature of '1>' is11 dtJds 11 = 11 i 11 /(1 - )."'s) = "'s/{1 - >"",s) .

2.8 The circle passes through 'Y(s) because 11 E - '1 11 = 11 K,lo , 11 = 1/I"'sl,which is the radius of the circle. It is tangent to '1 at this point becauseE- '1 = ...Ln s is perpendicular to the tangent t of '1. The curvature of the

K,

circle is the reciprocal of its radius, i.e. I"'sl, which is also the curvatureof '1.

2.9 The tangent vector of Eis t +...L( -"'st) - ~ns = -~ns so its arc-length~. K. K.

is u = J 11 E 11 ds = J~ds = Uo - ...L, where Uo is a constant. Hence,It. It.

the unit tangent vector of Eis -ns and its signed unit normal is t . Since3

-dns/du = "'st/(du/ds) = ~t, the signed curvature of E is ",~/ks .

Denoting d/dt by a dash, '1' = a(1 - cos t, sin t) so the arc-lengths of '1 is given by ds/dt = 2asin(t/2) and its unit tangent vectoris t = 'Y = (sin(t/2), cos(t/2)) . So n, = (- cos(t/2), sin(t/2)) andi = (dt/dt)/(ds/dt) = 4ash~(t72) (cos(t/2), - sin(t/2)) = -1/4asin(t/2)ns,so the signed curvature of'Y is -1/4asin(t/2). Its evolute is thereforeE= a(t-sin t, l-cos t) -4a sin(t/2)(- cos(t/2), sin(t/2)) = a(t+sin t, -1+cos t). Reparametrising by t = ?T+t, we get a(t-sin t, l-cos t)+a( -?T, -2),which is obtained from a reparametrisation of'Y by translating by the vector a( -?T, -2).

2.10 The free part of the string is tangent to '1 at 'Y(s) and has length l- s, hencethe stated formula for ,( s). The tangent vector of , is 'Y - 'Y + (l - s).:y= "'s(l - s)n, (a dot denotes d/ds). The arc-length v of , is given bydv/ds = "'s{l - s) so its unit tangent vector is n, and its signed unitnormal is -t. Now dns/dv =~1D.s = i-I t, so the signed curvature

1C..\~-8J -8

of s is 1/(l- s) .2.11 (i) With the notation in Exercise 2.9, the involute of Eis

dE 1,(u) = E+ (l- u)-d = '1 + -ns - (l- u)ns = '1 - (l- uo)ns,

u "'ssince u =Uo - ...L, so , is the parallel curve 'Y-(i-uo) .

K,

(ii) Using the results of Exercise 2.10, the evolute of s is

,+ (i- s)( -t) ='1+ (i- s)t - (i- s)t ='1.

2.12 Ir we take the fixed vector in Proposition 2.2 to be parallel to the line ofreflection, the effect of the reflection is to change cp to -cp, and hence to

change "'s to -"'S'2.13 Ir two unit-speed curves have the same non-zero curvature, their signed

curvatures are either the same or differ in sign . In the first case the curvesdiffer by a rigid motion by Theorem 2.1; in the latter case, applying areflection to one curve gives two curves with the same signed curvatureby Exercise 2.12, and these curves then differ by a rigid motion.

Solutions 287

2.14 (i) t = (~(1 + t)1/2, - Hl - t)1/2, ~) is a unit vector so 'Y is unit-speed;

i = (t(1 + t)-1/2, t(l- t)-1/2, 0), so ~ = 11 i 11 = 1/J8(1 - t2 )j n = ~i =~((1 - t)1/2, (1 + t)1/2,0)j b = t x n = (-Hl + t)1/2, ~(1- t)1/2, ~)j

b = (-t(l + t)-1/2 , -t(1 - t)-1/2, 0) so the torsion 7 = 1/J8(1 - t2 ) .

The equation il = -~t + 7n is easily checked.(ii) t = (- ~ sin t, - cos t, I sin t) is a unit vector so 'Y is unit-speed; i =(-~cost,sint,Icost), so ~ =:,11 i 11= I: n = ~i = (-~cost,sint,Icost);

b = t x n = (-1,0, -~), so b = 0 and 7 = O.2.15 Show that 7 = 0 or observe that x = 11t 2

, Y = t + 1, Z = l;t satisfyx - y - Z = O.

2.16 By Proposition 2.5, 'Y is a circle of radius 1/~ = 1 with centre 'Y +~n = (0,1,0) in the plane passing through (0,1,0) perpendicular tob = (-I, 0, -~), i.e. the plane 3x + 4z =O.

2.17 Let a = ~/(~2 + 72), b = 7/(~2 + 72). By Examples 2.1 and 2.4, thecircular helix with parameters a and b has curvature a/(a2 + b2 ) = ~ andtorsion b/(a2 + b2 ) = 7. By Theorem 2.3, every curve with curvature ~

and torsion 7 is obtained by applying a rigid motion to this helix.2.18 This follows from Proposition 2.3 since the numerator and denominator

of the expression in (11) are smooth functions of t.2.19 Let a dot denote d/dt . Then, 6=;y= xn, so the unit tangent vector of 6 is

T = n and its arc-Iength s satisfies ds/dt = n, Now dT/ds = il/(ds/dt) =~-l(-~t + 7b) = -t + ~b. Hence, the curvature of 6 is 11 - t + ~b 11 =(1 + ~)1/2 = JL, say. The principal normal of 6 is N = JL-l(-t + ~b)and its binormal is B = T x N == JL-l(b + Lt). The torsion T of 6 is. '"given by dB/ds = -TN, i.e, ~-lB = -TN. Computing the derivativesand equating coefficients of b gives T = (~f - 7K)/~(~2 + 72).

2.20 Differentiating t.a (= constant) gives n .a = O; since t, n, b are an orthonormal basis of R 3

, a = t cos0 + JLb for some scalar JLj since a is a unitvector, JL = ± sin 0; differentiating a = t cos0 ± b sin 0 gives 7 = ~ cot O.Conversely, if 7 = >'~, there exists 0 with >. = cot Oj differentiating showsthat a = t cos 0 + b sin 0 is a constant vector and t.a = cos 0 shows that() is the angle between t and a.

2.21 Differentiating ('Y - a) .('Y - a) = r2 repeatedly gives t .('Y - a) = Oj t.t +~n.('Y-a) = 0, and so n .('Y-a) = -I/~j n.t+ (-~t+7b).('Y-a) = K/~2 ,

and so b.('Y- a) = K/7~2j and finally b .t - 7n.('Y - a) = (K/7~2)", and so7 / ~ = (K/7~2)".

Conversely, if Eq. (20) holds, then p = -a(pa}, so (p2 + (pa)2)" = 2pp +2(pa)(pa)" = 0, hence p2 + (pa)2 is a constant, say r 2 (where r > 0). Leta = 'Y+pn+ pabj then ä.= t+pn+ p(-~t+7b) + (pa)"b+ (pa)(-7n) = 0using Eq. (20); so ais a constant vector and 11 'Y - a 11 2 =p2 + (pa)2 = r2,


hence -y is contained in the sphere with centre a and radius r .2.22 Let Aij = v i .v] , The vectors VI, V2 , Va are orthonormal if and only if

Aij = &i j (= 1 if i = j and = 0 if i =I j) . So it is enough to prove thatAij = 8i j for all values of 8 given that it holds for 8 = 80 . Differentiatingvi .v, gives

a).ij = 'L)aikAkj + ajkAik).

k=l

Now Aij = 8i j is a solution of this system of differential equations becausea ij + aji = O. But the theory of ordinary differential equations teIls usthat there is a unique solution with given values when 8 = 80 .

Chapter 3

3.1 If;Y is obtained from -y by a translation, then .y = l' so by Eqs. (1) and (3)the length and area of;Y are the same as those of -y. If;Y is obtained from -yby a rotation by an angle () about the origin, then x= x cos ()- Ysin (), fj =x sin ()+Y cos (). This implies that &;2 +t/ =±2 + il and Xii - fj&; = xiJ - y±,so Eqs. (1) and (3) again show that the length and area are unchanged.

3.2 -y(t') = -y(t) <===} cost' = cost and sint' = sint <===} t' - t is a multiple of27r, so -y is simple closed with period 27r. Taking x = a cos t, y = bsin t inEq. (3) gives the area as ~ I: 1r

abdt = nab.3.3 l' =(- sin t - 2 sin 2t, cos t + 2 cos 2t) so 11 l' 11 =V5 + 4 cos t, which is never

zero, so -y is regular. -y(t+ 27r) = -y(t) is obvious. If -y is simple closed withperiod a, then -y(O) = -y(a) implies (1 + 2cosa) cosa = 3 so cosa = 1 anda = 27r. But -y(2rr/3) = -y(47r/3) and 47r/3 - 2rr/3 is not a multiple of 27r .

3.4 Differentiating -y(t+ a) = -y(t) gives t(t + a) = t(a); rotating anti-clockwiseby 7r/2 gives ns(t+a) = ns(t); differentiating again gives Ks(t+a)ns(t+a) =Ks(t)ns(t), so Ks(t + a) = Ks(t).

3.5 By Exercise 3.2 and the isoperimetric inequality, the length i of the ellipsesatisfies i ~ v'47r x 7rab = 27r../äb, with equality if and only if the ellipseis a circle, i.e, a = b. But parametrising the ellipse as in Exercise 3.2, itslength is

r21r r21r

Jo

11 l' 11 dt = Jo

Ja2 sin2 t +v cos2 t dt.

3.6 Let (Xl, YI) and (X2, Y2) be points in the interior of the ellipse, so that2 2

~ + ~ < 1 for i = 1,2. A point of the line segment joining the two pointsis (txI +(1- t)X2,tYI +(1- t)Y2) for some 0 $ t $ 1. This is in the interior

Solutions 289

of the ellipse because

(txI + (1 - t)X2)2 (tYI + (1 - t)Y2)2a2 + b2

=t2 (x~ + Y~) + (1- t)2 (x~ + Y~) + 2t(1- t) (X IX2 + YIY2)a2 b2 a2 v a2 b2

(x2 y2 x2 y2)< t2 + (1 - t)2 + t(l - t) ---!.. + -.!. + .2 + 2a2 b2 a2 b2

< e + (1 - t)2 + 2t(1 - t) = 1.

3.7 If i(t) is a reparametrisation of 'Y(t), the vertices of'Y (resp. i) are given bydKs/dt = 0 (resp . dKs/dt = 0). Since dKs/dt = (dKs/dt)(dt/dt) and sineedt/dt is never zero, 'Y and i have the same vertices.

3.8 'Y = (- sin t - 2 sin 2t,cos t + 2cos 2t) and 11 'Y 11 = y"'5-:-+-'4'-e-os"""7t, so the angle<p between 'Y and the z-axis is given by

- sin t - 2sin 2t . eos t + 2 eos 2teosif) = , Sln'f) = .

T V5 + 4eost T V5 + 4eost

Differentiating the seeond equation gives <p eos <p = - sin t(24 eos2 t +42cost + 9)/(5 + 4cost)3/2, so <p = sint(24eos2t + 42eost + 9)/(5 +4cost)(sint + 2sin 2t ) = (9 + 6eost)/(5 + 4eost). Henee, if s is the arclength of 'Y, Ks = d<p/ds = (d<p/dt)/(ds/dt) = (9 + 6 eos t)/(5 + 4 eos t)3/2,so Ks = 12sin t(2 + eos t)/(5 + 4 COSt)5/2. This vanishes at only two pointsof the eurve, where t = 0 and t = 1r.

Chapter 4

4.1 Let U be an open disc in R 2 and S = {(x, Y, z) E R 3 I (x, y) E U, z = O}.If W = {(x , y, z) E R 3 I (x, y) EU}, then W is an open subset of R 3

,

and Sn W is homeomorphic to U by (x , y, 0) t-t (x, y) . So S is a surface.4.2 Let U = {(u,v) E R 2 10< u2+v2 < 11"2}, let r = vu2 +v2, and define

a : U -+ R 3 by u(u,v) = (;, ~,tan(r - I».4.3 The image of u~ is the intersection of the sphere with the open set ±x > 0

in R 3, and its inverse is the projection (x, y, z) t-t (y, z). Similarly for u~

and u±. A point of the sphere not in the image of any of the six patcheswould have to have x , y and z all zero, which is impossible.

4.4 Multiplying the two equations gives (X2_ Z2)sin 8 eosB = (1_ y2)sin 8 eos 8,so x2+y2 - z2 =1 unless cos8 =0 or sin8 =0; if cos8 = 0, then x = -zand y = 1 and if sin 8 =0 then x =z and y =-1, and both ofthese linesare also eontained in the surface.


The given line L o passes through (sin 28, - cos 28,0) and is parallel to thevector (cos 28, sin 28, 1) j it follows that we get all of the lines by takingo ::; 8 < tt , Let (x , y , z) be a point of the surface; if x i- z , let 8 be suchthat cot8 = (1 - y)/(x - z}; then (x,y, z) is on LOi similarly if xi- - z.The only remaining cases are the points (0,0, ±1), which lie on the linesL 7r / 2 and u;With the notation of Exercise 4.2, define (1 : U -+ R 3 by (1(u ,v) =(sin 28, - cos 28,0) + t(cos 28, sin 28,1) , where t = tan(r - ~), cos 8 = u/rand sin 8 = u]«, This is a surface patch which, by the preceding paragraph,covers the whole surface.Let M <p be the line

(x - z) cos e = (1 + y) sin <p, (x + z) sin <p = (1 - y) cos <po

By the same argument as above, M<p is contained in the surface andevery point of the surface lies on some M<p with 0 ::; sp < 'Tr. Ir 8 +<p is not a multiple of 1r, the lines L o and M <p intersect in the point( cos(O-<p) sin 0- '1' cos(O+<p») r h LI • h 0 < LI here i 1

sin( O+<p) 'sin(O+<p ' sin(O+ <p ) i ror eac U wit _ U < rr, t ere IS exact yone <p with 0 ::; <p < 1r such that 8 + sp is a multiple of n , and the lines Loand M <p do not intersect. Ir (x, y, z) lies on both L o and L <p , with 8 i- sp ,then (1 - y) tan 8 = (1 - y) tan <p and (1 + y) cot 8 = (1 - y) cot <p, whichgives both y = 1 and y = -1 (the case in which 8 = 0 and sp = 1r/2 , orvice versa, has to be treated separately, but the conclusion is the same).This shows that L o and L <p do not intersect; similarly, Mo and M <p do notintersect.

4.5 Ir the sphere S could be covered by a single surface patch (1 : U -+ R 3, then

S would be homeomorphic to the open subset U of R 2• As S is a closed

and bounded subset of R 3, it is compact. Hence, U would be compact,

and hence closed. But, since R 2 is connected, the only non-empty subsetof R 2 that is both open and closed is R 2 itself, and this is not compactas it is not bounded.

4.6 (1 is obviously smooth and (1u X a; = (-lu, - Iv, 1) is nowhere zero, so (1

is regular.4.7 (1± is a special case of Exercise 4.6, with I = ±Vl - u2 - v2 (VI - u2 - v2

is smooth because 1 - u2 - v2 > 0 if (u, v) E U) i similarly for the otherpatches. The transition map from (1,+- to (1~, for example, is tP(ü ,v) =(u,v) , where (1~(ü,v) = (1,+-(u ,v) j so u = Vl- ü2 - v2, V = v, and this issmooth since 1- ü2 - v2 > 0 if (ü,v) EU.

4.8 (1 is a smooth map on the open set R = {(r,8) E R 2 Ir> O} , andits image is contained in the surface because cosh28 - sinh28 = 1; and(1r X (10 = (-2r2cosh 8, 2r2sinh 8, r), which is never zero on R.Exercise 4.6 gives the parametrisation Ö'(u, v) = (u, v, u2- v2), defined on

Solutions 291

the open set U = {(u, v) E R 2 I u2 - v2 > O}. This is a reparametrisationof a via the reparametrisation map (u, v) 1--7 (r,O), where r = vu2 - v2 ,

0= cosh-1(u/VU2 - v2) (this is smooth because u2 - v2 > 0).For the part with z < 0, we can take u(r,O) = (r sinhO, r cosh 0, _r2 )

defined on the open set R; and Ö'(u,v) = (u,v,u2 - v2 ) defined on theopen set V = {(u,v) E R 2 I u2 - v2 < O}.

4.9 This is similar to Example 4.5, but using the 'latitude longitude' patchu(O,cp) = (acosOcoscp,bcosOsincp ,csinO). Alternatively, one can use

Theorem 4.1, noting that if I(x , y, z) =~+~+~ -1, then (Ix , I y , Iz) =(2x/a2,2y/b2,2z/c2) vanishes only at the origin , and hence at no point ofthe ellipsoid.

4.10 A typical point on the circle C has coordinates (a + bcos0,0, bsin 0); rotating this about the z-axis through an angle sp gives the point u(O, cp)jthe whole of the torus is covered by the four patches obtained by taking (0, cp) to lie in one of the following open sets : (i) 0 < 0 < 27r, 0 <sp < 27r, (ii) 0< 0 < 27r, -7r < cp < 7r, (iii) -7r < 0 < 7r,0 < cp < 27r,(iv) -7r < 0 < 7r, -7r < sp < rr. Each patch is regular becauseUe x ul{> = -b(a + bcosO)(cosOcoscp,cosOsincp, 1) is never zero (sincea + bcosO ~ a - b > 0).Let u(O,cp) = (x,y,z) . Then, x2 + y2 + Z2 + a2 - b2 = 2a(a + bcosO), so(x2+y2+ z2+ a2_b2)2 = 4a2(a+bcosO)2 = 4a2(x2+y2) . Let I(x,y ,z) bethe left-hand side minus the right-hand side; then, Ix = 4x(x2+ y2 + z2 a2 - b2), I y = 4y(x2+ y2+ z2 - a2- b2), Iz = 4z(x2+ y2+ z2+a2 - b2); ifI z = 0 then z = 0 since x2+y2+Z2 +a2- b2 > 0 everywhere on the torus;if Ix = I y = 0 too, then since the origin is not on the torus, we must havex2 + y2 = a2 + b2, but then substituting into the equation of the torusgives (2a2)2 = 4a2(a2+b2), a contradiction. So (Ix, I y, fz) is nowhere zeroon the torus, which is therefore a smooth surface by Theorem 4.1.

4.11 If S is covered by a single surface patch a : U ~ R 3, then I : S ~ R is

smooth if and only if I 0 u : U ~ R is smooth. We must check that, ifÖ' : (j ~ R 3 is another patch covering S, then 10 Ö' is smooth if and onlyif I 0 a is smooth. This is true because I 0 Ö' = (I 0 u) 0 1/>, where I/> is thetransition map from a to Ö', and both I/> and 1/>-1 are smooth.

4.12 If Ö' =u+a, where a is a constant vector, then Ö' is smooth if a is smooth,and Ö'u =U u , Ö'v =U v, so Ö' is regular if a is regular.If Ö' = A 0 a, where A is a linear transformation of R 3 , then 0' is smoothif a is smooth, and O'u =A(uu),O'v =A(uv), so, if Ais invertible, O'u andO'v are linearly independent if Uu and U v are linearly independent.

4.13 (i) At (1,1,0), Uu = (1,0,2) , U v = (0,1 , -2) so Uu x U v = (-2,2,1) andthe tangent plane is -2x + 2y + z = O.(ii) At (1,0,1), where r = 1,0 = 0, Ur = (1,0,2), Ue = (0,1,2) so Ur XUe =


(-2, -2, 1) and the equation of the tangent plane is -2x - 2y + z = 0.4.14 Suppose the centre of the propeller is initially at the origin. At time t, the

cent re is at (0,0, at) where o is the speed of the aeroplane. If the propelleris initially along the z-axis, the point initially at (v, 0, 0) is therefore atthe point (v coswt,v sin wt, at) at time t , where w is the angular velocityof the propeller. Let U = wt, ,X = ccli».fTu = (-vsinu,vcosu,'x) , fTv = (cosu,sinu,O), so the standard unit normal is N = (,X2 +v2)-1/2(-'xsinu,'xcosu, -v) . IfO is the angle betweenN and the z-axis, cosO = -V/(,X2 +v2)1/ 2 and hence cotO = ±v/'x, whilethe distance from the z-axis is v.

4.15 Let Ö"(u, u) be a reparametrisation of a . Then,

8u_ 8u_ 8u_ 8u_fTu = 8u fTü + 8u fTu, o; = 8v fTü + 8v fTu,

so a.; and a; are linear combinations of Ö"Ü and Ö"U. Hence, any linearcombination of fTu and fTv is a linear combination of Ö"Ü and Ö"U . Theconverse is also true since o is a reparametrisation of Ö".

4.16 If 'Y(t) = (x(t), y(t), z(t)) is a curve in 5, differentiating I(x(t), y(t), z(t)) =°gives Ixx+lyiJ+Izz =0, i.e. V I."y =0, showing that V I is perpendicularto the tangent vector of every curve in 5, and hence to the tangent planeof 5 . Since 5 has a canonical (smooth) choice of unit normal V I /11 VI 11at each point , it is orientable.

4.17 See the proof of Proposition 11.1 for the first part. By the argument inExercise 4.16, VF."y(O) = !tF(-y(t))lt=o, so VsF- VF is perpendicular tothe tangent plane of 5 at P. This proves that Vs F is the perpendicularprojection of VF onto the tangent plane . If F has a local maximum orminimum at P, then t ~ F(-y(t)) has a local maximum or minimum att =°for all curves 'Y on 5 with 'Y(O) = P; hence, VsF = 0 at P, so VFis perpendicular to the tangent plane of 5 at P, and hence is parallel toVI by Exercise 4.16. So VF = ,XVI for some scalar X.

4.18 From Example 4.13, the surface can be parametrised by fT(u,v) =(cosh u cosv, cosh u sinv, u), with u E Rand -11" < v< 11" or °< v < 211".

4.19 11 fT(u,v) 11 2 = sech2u(cos2 v+sin2 v) +tanh2u =sech2u+tanh2u = 1, so fTparametrises part of the unit sphere; (1 is c1early smooth; and ou X av =-sech2u(1(u, v) is never zero, so a is regular. Meridians correspond to theparameter curves v = constant, and parallels to the curves u = constant.

4.20 The vector au is tangent to the meridians, so a unit-speed curve y is a loxodrome if "y.fTu/ 11 au 11 = cosa , which gives u= cosh u cos o ; since 'Y is unitspeed, "y = (-u sech u tanh u cosv - i.J sech u sin v, -u sech u tanh u sin v +i.J sech u cosv, Ü sech2u) is a unit vector; this gives ü2 + i.J2 = cosh'' U, soi.J = ± cosh u sin o. The correponding curve in the uv-plane is given by

Solutions 293

dv/ du = iJ /iJ. = ± tan o , and so is a straight line v = ±u tan a + c, wherecis a constant.

4.21 The point at a distance v from the z-axis on the ruling through (0,0, u)has position vector given by O'(u, v ) = (O ,O,u) + v(cosB(u) ,sinB(u) ,O) =(vcosB(u) ,vsinB(u),u) ; O'u x O'v = (-sinB ,-cosB,-vdB/du) is clearlynever zero, so 0' is regular.

4.22 (i) i'.a =°so i' is contained in the plane perpendicular to a and passingthrough the origin; (ii) simple algebra; (iii) ii is clearly a smooth function

of (u, v) and the jacobian matrix of the map (u, v) ~ (u, ii) is ('Y~a ~),where a dot denotes d/du; this matrix is invertible so i' is a reparametrisation of 'Y.

4.23 (i) (a cos u cosv , bcos u sin v, csin u) (cf. Exercise 4.9); (ii) see Exercise 4.4 ;

(iii) (u, v, ± /1 + ~ + ~)j (iv), (v) see Exercise 4.6j (vi) see Example 4.3 jV p q

(vii) (p cos u, qcos u, v); (viii) (±pcosh u, qsinh u, u): (ix) (u ,U2/ p2,v) ; (x)(0, u , v); (xi) (±p,u, v).

4.24 (a) Types (viij-fxi); (b) type (vi) ; (c) types (ii) (see Exercise 4.4), (v) (seeExercise 4.25) and (vij-Ix); (d) type (i) if p2,q2 and r 2 are not distinct ,types (ii) , (iii) , (iv), (vi) and (vii) if p2 = q2, and type (x).

4.25 z = (~- ~) (~ + ~) = uv , x = ~p(u + v), Y = ~q(v - u) , so

a parametrisation is O'(u,v) = (~p(u + v) , ~q(v - u),uv); O'u x O'v =(-~q(u + v), ~p(v - u) ,pq) is never zero so 0' is regular. For a fixed valueof u, O'(u,v) = (~pu , -~qu,O) +v(~p, ~q,u) is a straight line; similarly fora fixed value of Vj hence the hyperbolic paraboloid is the union of each oftwo families of straight lines.

4.26 Substituting the components (x, y, z) of 'Y(t) = a + bt into the equationof the quadric gives a quadratic equation for t; if the quadric containsthree points on the line, this quadratic equation has three roots, hence isidentically zero, so the quadric contains the whole line.Take three points on each of the given lines; substituting the coordinatesof these nine points into the equation of the quadric gives a system ofnine homogeneous linear equations for the ten coefficients al, ... , c of thequadric; such a system always has a non-trivial solution. By the first part,the resulting quadric contains all three lines.

4.27 Let LI, L 2 , L 3 be three lines from the first family; by Exercise 4.26, thereis a quadric Q containing all three Iines; all but finitely many lines ofthe second family intersect each of the three lines; if L' is such a line, Qcontains three points of L', and hence the whole of L' by Exercise 4.26jso Q contains all but finitely many lines of the second family ; since anyquadric is a closed subset of R 3

, Q must contain all the lines of the second

294 Efementary Differential Geometry

family, and hence must contain S .4.28 Both parts are geometrically obvious.4.29 Let (a, b,c) be a point of R 3 with a and b non-zero . Then Ft(a, b,c) -t 00

as t -t 00 and as t approaches p2 and q2 from the left; and Ft(a , b, c) -t

-00 as t -t -00 and as t approaches p2 and q2 from the right. Fromthis and the fact that Ft(a, b, c) = 0 is equivalent to a cubic equationfor t, it follows that there exist unique numbers u, v, w with u < p2,p2 < V < q2 and q2 < w such that Ft(a, b, c) =0 when t =u, vor w. Thesurfaces Fu(x, y, z) = 0 and Fw(x, y, z) = 0 are elliptic paraboloids andFv(x , y , z) = 0 is a hyperbolic paraboloid, and we have shown that thereis one surface of each type passing through each point (a, b,c).To parametrise these surfaces, write Ft(x, y, z) = 0 as the cubic equationX2(q2 _t)+y2(P2 -t) -2z(p2_t)(q2 -t)+t(P2 -t)(q2 -t) = 0, and note thatthe left-hand side must be equal to (t - u)(t - v)(t - w); putting t =p2,q2

and then equating coefficients of t 2 (say) zives x = ±..j(p2_U)(p2_V)(p2_W)O' q2_p2'

Y =±..j(q2_U)(q2_ v)(q2_w) z = !(u + v + w _ p2 _ q2).p2_q'i , 2

4.30 Let F : W -t V be the smooth bijective map with smooth inverse F -l :V -t W constructed in the proof of Proposition 4.1. Then, (u(t), v(t)) =F-1 (-y(t)) is smooth.

4.31 Suppose, for example, that fy :f: 0 at (xo, Yo). Let F(x, y) = (x,J(x, y)) ;

then F is smooth and its jacobian matrix (~ ~:) is invertible at

(xo, Yo) . By the inverse function theorem, F has a smooth inverse G defined on an open subset of R 2 containing F(xo, Yo) = (xo,0), and G mustbe of the form G(x,z) = (x ,g(x,z)) for some smooth function g. Then-y(t) = (t,g(t,O)) is a parametrisation of the level curve f(x ,y) = 0 containing (xo, Yo) .

The matrix (fx fy fz) has rank 2 everywhere ; suppose that, at somegx gy gz

point (xo, Yo, zo) on the level curve, the 2 x 2 submatrix (fY fz) isgy gz

invertible. Define F(X'y,z();= (Xof(tiy),z),g(x,y,z)) ; then Fis smooth

and its jacobian matrix fx fy fz is invertible at (XO,yo,zo). Letgx gy gz

G(x, u, v) = (x, cp(x, u, v), .,p(x, u, v)) be the smooth inverse of F definednear (xo, 0, 0). Then -y(t) = (t, cp(t, 0, 0), .,p(t,0, 0)) is a parametrisation ofthe level curve f(x, y, z) =g(x, y, z) =0 containing (xo, s», zo).

Solutions

Chapter 5

295

5.1 (i) Quadric cone; we have Uu = (cosh usinh v, cosh u cosh v , cosh u) , a; =(sinhu cosh v, sinhusinhv, 0), and so we get 11 Uu 11 2 = 2cosh2u, uu.uv =2sinhucoshusinhvcoshv, 11 a; 11 2 = sinh2u, and the first fundamentalform is 2 cosh'' U du2 + 4 sinh u cosh u sinh v cosh v dudv + sinh2 u dv2.

(ii) Paraboloid of revolution; (2 + 4u2) du2 + 8uv dudv + (2 + 4v2) dv2.(iii) Hyperbolic cylinder; (cosh2 u + sinh2 u) du2 + dv2.

(iv) Paraboloid of revolution; (1 + 4u2)du2 + 8uv dudv + (1 + 4v2) dv2.5.2 The first fundamental form is 2 du2+ u2 dv2, so the length of the curve is

Jo'/l" (2u2 + U 2V2)1/2dt = Jo'/l" (2).2 e2>.t + e2>.t)1/2 dt = (2).2~l) lI2 (e>.t - 1).

5.3 Applying a translation to a surface patch o does not change Uu or Uv' HAis a rotation about the origin, (Au)u = A(uu), (Au)v = A(uv), and Apreserves dot products (A(p) .A(q) = p .q for all vectors p ,q E R 3

) .

5 4 B th hai I - - &u &v - - &u &v hi h .. y e c am rue, Uü - UU&ü +UV&ü' Ujj - UU8jj +UV8jj ' w lC grves

E- - - - _E(&U)2 2F 8u 8v G(&V)2 S' '1 . J: F- d_ - uü .uü - 8ü + 8ü 8ü + 8ü' Iml ar expreSSlOns lor anG can be found; multiplying out the matrices shows that these formulasare equivalent to the matrix equation in the question.

5.5 The map is u(u, v) I--t (uV2 cos 0 'uV2sin 0 ,0) = Ö'(U, v), say. The image of this map is the sector of the xy-plane whose polar coordinates (T,9)satisfy 0 < 9 < 1rV2.The first fundamental form of u is 2 du2+u2dv2 (Ex-

ercise 5.2); Ö'u = (V2cos ~, V2sin~,0), Ö'v = (-usin ~, ucos ~, 0) ,so 11 Ö'u 11 2= 2, uU'uv = 0, 11 a; 11 2= u2 and the first fundamental form ofÖ' is also 2 du2 + u2 dv2.

5.6 No: the part of the ruling (t, 0, t) with 1 ~ t ~ 2 (say) has length .,fi andis mapped to the straight line segment (t, 0, 0) with 1 ~ t ~ 2, which haslength 1.

5.7 For the generalised cylinder, Example 5.3 shows that the first fundamental


form is du2 + dv2, so O'(u, v) t-t (u, v,0) is an isometry from the cylinderto part of the xy-plane. For the generalised cone, Example 5.4 shows thatthe first fundamental form is v2du2 + dv2. This is the same as the firstfundamental form of Ü (.12' uv'2), where ü is as in Exercise 5.5. Since üparametrises part of the plane, the generalised cone is isometrie to partof the plane.

5.8 A straightforward calculation shows that the first fundamental form of O'tis coslr' u(du2+dv2)j in particular, it is independent of t . Hence, O'(u, v) t-t

O't (u, v) is an isometry for all t, Taking t = 'Ir/2 gives the isometry fromthe catenoid to the helicoid; under this map, the paralleIs u = constanton the catenoid go to circular helices on the helicoid, and the meridiansv = constant go to the rulings of the helicoid .

5.9 If the first fundamental forms of two surfaces are equal, they are certainlyproportional, so any isometry is a conformal map . Stereographie projection is a conformal map from the unit sphere to the plane, but it is notan isometry since >. ::11 (see Example 5.7).

5.10 The vector O'u is tangent to the rulings ofthe cone, so the angle () betweenthe curve and the rulings is given by cos() = "/'O'u/ 11 "/ 1II1 O'u 11. l.From')'(t) = (eAtcost,eAtsint ,eAt) and O'u = (cost,sint,l) at ')'(t), we getcos() = J2>.2/(2).2 + 1), whieh is independent oft.

5.11 The first fundamental form is sech2u(du2 + dv2 ) .

5.12 The first fundamental form is (1 + P)du2 + u2dv2, where a dot denotesd/du . So 0' is conformal if and only if i = ±vu2 - 1, i.e, if and only iff(u) = ±(tuvu2 - 1 - t cosh"? u) + c, where c is a constant.

5.13 The first fundamental form is (1+ 2v"/.6+v26.6)du2+ 2"/.6dudv +dv2. So0' is conformal if and only if 1 + 2v"t.6+ v26.6 = 1 and "/.6 = 0 for all u, Vjthe first condition gives 6= 0, so 6 is constant, and the second conditionthen says that ')'.6 is constant, say equal to d. Thus, 0' is conformal if andonly if 6 is constant and ')' is contained in a plane r .6 = d. In this case, 0'

is a generalised cylinder.

5.14 0' is conformal if and only if I~ + g~ = f; + g; and lulv + gugv =O. Letz = lu +igu, w = Iv +igvj then 0' is conformal if and only if zz = ww andzw + zw = 0, where the bar denotes complex conjugate; if z = 0, thenw = 0 and all four equations are certainly satisfied; if z ::I 0, the equationsgive z2 = _w2 , so Z = ±iwj these are easily seen to be equivalent tothe Cauchy-Riemann equations if the sign is +, and to the 'ant i-CauchyRiemann' equations if the sign is - .

5.15 Parametrise the paraboloid by O'(u, v) = (u, v, u2+v2), its first fundamental form is (1 + 4u2)du2 + 8uv dudv + (1 + 4v2)dv2. Hence, the requiredarea is JJ JI + 4(u2+ v2)dudv, taken over the disc u2 + v2 < 1. Let

u =r sin B, v =r cosBj then the area is 2'1r J; vI + 4r2r dr = *(53/

2- 1).

Solutions 297

This is less than the area 21T of the hemisphere .5.16 Parametrize the surface by O'(u ,v) = (p(u) cosv,p(u) sinv, a(u)) , where

'Y(u) = (p(u), 0, a(u)) . By Example 6.2, the first fundamental form isdu2 + p(U)2 dv2, so the area is JJ p(u) dudv = 21T J p(u) du.(i) Take p(u) = cosu, a(u) = sinu, with -1T/2 ~ u ~ 1T/2j so

21T J~~~2 COS u du = 41T is the area.(ii) For the torus, the profile curve is 'Y(O) = (a + bcosO,O,bsinO),but this is not unit-speed; a unit-speed reparametrisation is .y(u) =(a+bcosI,O ,bsinI) with °~ u ~ 21Tb. So 21TJ:7I"b(a+bcosI) du =41T2ab is the area.

5.17 0' is the tube swept out by a circle of radius a in aplane perpendicular to'Y as its centre moves along 'Y. 0'a = (1 - IW cosO)t - Ta sin On+ Tacos Ob,0'9 = -asinOn+acosOb, givingO'aX0'9 = -a(1-KacosO)(cosOn+sinOb);this is never zero since na < 1 implies that 1 - sa cos0 >°for all O.The first fundamental form is «1 - KacosO)2 + T2a2)ds2 + 2Ta2dsdO +a2d02, so the area is J;l J0271" a(l - na cosO)dsdO = 21Ta(SI - so).

0 2 25.18 If EI = E2,F1 = F2,G1 = G2, then E1G1 - F1 = E2G2 - F2, so anyisometry is equiareal. The map f in Archimedes's theorem is equiarealbut not an isometry (as EI ::j:. ~, for example).

5.19 If EI = >'~,F1 = >.F2,GI = >'G2, and if E1G1 - Fr =~G2 - F:j, then>.2 =1 so >. = 1 (since >. > 0).

5.20 By Theorem 5.5, the sum of the angles of the tri angle is 1T + A/R2 ,

where A is its area and R is the radius of the earth, and so is ~

1T + (7500000)/(6500)2 = 1T + 136°9 radians. Hence, at least one angleof the tri angle must be at least one third of this, i.e, 1T + 116°9 radians.

5.21 3F = 2E because every face has three edges and every edge is an edgeof exactly two faces. The sum of the angles around any vertex is 21T, sothe sum of the angles of all the triangles is 21TV; on the other hand, byTheorem 5.5, the sum of the angles of any triangle is 1T plus its area, sosince there are F triangles and the sum of all their areas is 41T (the areaof the sphere), the sum of all the angles is 1TF + 41T. Hence, 2V = F + 4.Then, V - E + F = 2 + ~F - E + F = 2 + ~ (3F - 2E) = 2.

5.22 Let 0' : U -+ R 3. Then, f is equiareal if and only if

!L(EI GI - Ff)1/2 dudv =!L(E2G2 - Fi)1/2 dudv

for all regions R ~ U. This holds if and only if the two integrands areequal everywhere, i.e. if and only if EI GI - Fr = ~G2 - Fi .

298

Chapter 6

Elementary Differential Geometry

6.1 (Tu = (1,0, 2u), a; = (0,1, 2v), so N = .>.( -2u, -2v, 1), where .>. = (1 +4u2+4v2

) - 1/ 2 j (Tuu = (0,0,2), (Tuv = 0, (Tvv = (0,0,2), so L = 2'>', M = 0,N = 2'>', and the second fundamental form is 2'>'(du2 + dv2 ) .

6.2 (Tu .Nu = -(Tuu.N (since (Tu.N = 0), so Nu .(Tu = O; similarly, Nu.(Tv =Nv·(Tu = Nv .(Tv = O; hence, Nu and N, are perpendicular to both (Tuand (Tv, and so are parallel to N . On the other hand, Nu and N v areperpendicular to N since N is a unit vector. Thus, Nu = N, = 0, andhence N is constant. Then, ((T.N)u = (Tu.N = 0, and similarly ((T.N)v = 0,so (T.N is constant, say equal to d, and then (T is part ofthe plane r.N = d.

6.3 From Section 4.3, N = ±N, the sign being that of det(J). From Ö'ü =8u + 8v - _ 8u + 8v t(Tu 8ü a; 8ü' (Tii - (Tu 8ii a; 8ii' we ge

_ 02u 02V (OU)2 OU ov (OV)2(Tüü = (Tu ou2 + (Tv ou 2 +(Tuu ou +2(Tuvou ou +(Tvv ou

So

L = ± (L (OU)2 2MoUOV N(OV)2)OU + ouou+ OU

since (Tu .N = (Tv.N = 0. This, together with similar formulas for !VI andN, are equivalent to the matrix equation in the question.

6.4 Applying a translation to a surface patch (T does not change (Tu and (Tv , andhence does not change N, (Tuu, (Tuv or (Tvv, and hence does not change thesecond fundamental form. A rotation A about the origin has the followingeffect: (Tu ~ A((Tu), (Tu ~ A((Tv) and hence N ~ A(N) , (Tuu ~ A((Tuu),(Tuv ~ A((Tuv), (Tvv ~ A((Tvv)j since A(p) .A(q) = p.q for any vectorsp, q E R 3

, the second fundamental form is again unchanged.6.5 By Exercise 6.1, the second fundamental form ofthe paraboloid is 2(du2 +

dv2)/Jl + 4u2 + 4v2 j so

"'n =2((- sin t)2 + cos2 t)/v'1 + 4cos2 t + 4sin2 t =2/../5.

6.6 ",2 = "'~ + "'~ = 0, so '" =°and 'Y is part of a straight line.6.7 Let 'Y be a unit-speed curve on the sphere of centre a and radius r . Then,

('Y - a).('Y - a) = r 2 j differentiating gives 'Y.('Y - a) = 0, so ;Y.('Y - a) =-1'.1' = -1. At the point 'Y(t), the unit normal of the sphere is N =±~('Y(t) - a), so "'n = ;Y.N = ±~;Y.('Y - a) = =F~.

6.8 Ir the sphere has radius R, the parallel with latitude () has radius r =R cos (}j if P is a point of this circle, its principal normal at P is parallel tothe line through P perpendicular to the z-axis, while the unit normal tothe sphere is parallel to the line through P and the centre of the sphere.

Solutions 299

The angle t/J in Eq. (8) is therefore equal to () or 'Ir - () so K g =±~ sin () =±*tan (). Note that this is zero if and only if the parallel is a great circle.

6.9 The unit normal is N = (-gcosv, -gsin v, j) , where a dot denotes d/du.On a meridian v = constant, we can use u as the parameter; since (Tu =(j cosv, j sin v, g) is a unit vector, u is a unit-speed parameter on themeridian and

jcosv jsinv jjK g = (Tuu .(N x (Tu) = -gcosv -gsinv j =0.

jcosv jsinv 9On a parallel u = constant, we can use v as a parameter, but a; =(- f sin v, f cosv, 0) is not a unit vector ; the arc-length s is given byds/dv = 11 (Tv 11 = f(u), so s = f(u)v (we can take the arbitrary constant to be zero). Then,

1 ( 1) 1 - f cos v - f sin v 0 jK g = f(u) 2(TVV ' N X f(u)(Tv = f(u)3 -gcosv -gsinv j = 7'

- f sin v f cosv 0

KIN2 - K2Nl =K«(N1.n)N2 - (N2.n)N1) =K(N1 x N 2) x n .

Taking the squared length of each side, we get

Ki + K~ - 2KIK2Nl.N2 = K211 (NI x N 2) x n 11

2 .

Now, N 1.N2 = coso ; 'Y is perpendicular to NI and N 2, so NI X N 2 isparallel to 'Y, hence perpendicular to n; hence,

11 (NI x N 2) x nll = 11 NI X N2 1111 n 11 = sin o .

6.11 N .n =cost/J, N.t = 0, so N .b = sint/Jj hence, N = ncost/J + b sin e andB = t x (n cose + b sin e) = b cose - n sin e. Hence,

N =ncos t/J +bsin t/J + .,p(-n sin t/J + b cos t/J)

= (-Kt + rb) cos t/J - nr sin t/J + .,p(-n sin t/J + b cos t/J)

= -K cos t/Jt+ (r + .,p)(bcos t/J - n sin t/J)

=-Knt + rgB.

The formula for B is proved similarly. Since {t, N, B} is a right-handedorthonormal basis of R 3

, Exercise 2.22 shows that the matrix expressingi, N,B in terms of t, N, B is skew-symmetric, hence the formula for i.

6.12 A straight line has a unit-speed parametrisation -y(t) = P + qt (with q aunit vector), so ;y = 0 and hence Kn = ;Y.N = O.In general, Kn = 0 {::::::} ;y is perpendicular to N {::::::} N is perpendicularto n {::::::} N is parallel to b (since N is perpendicular to t).


6.13 The second fundamental form is (-du2 + u2dv2)/uV1 + u2, so a curve-y(t) = O'(u(t),v(t)) is asymptotic if and only if -u2 + u2iJ2 = 0, i.e,dofdu = iJju = ±1/u, so lnu = ±(v+ c), where c is a constant.

6.14 By Exercise 6.12, b is parallel to N , so b = ±N; then, B = t x N = =fn.

Hence, B = =fit = =f(-Kt+rb) = ±Kt-rNj comparing with the formulafor B in Exercise 6.11 shows that rg = r (and Kn =±K).

6.15 For the helicoid 0'(u, v) = (v cos u, v sin u, ,xu), the first and second fundamental forms are (,X2 + v2)du2+ dv2 and 2'xdudv/J,X2 + v2, respectively.Hence , the principal curvatures are the roots of

I- K(,X2 + v2) ,,;}+v2!_

x -0,~ -K

i.e. ±,X/(,X2 + v2 ) .

For the catenoid O'(u,v) = (coshucosv,coshusinv,u) , the first and second fundamental forms are cosh'' u(du2+dv2) and -du2+dv2, respectively.Hence , the principal curvatures are the roots of

I- I - K COSh

2u 0 I

o 1 - Kcosh2u '

i.e, K = ±sech2u.6.16 Let s be arc-length along -y, and denote df ds by a dash. Then, by Propo

sition 6.1,

_L,2 2M" N,2 _ Lu2 + 2MuiJ + NiP _ Lu2 + 2MüiJ + NiJ2

K n - U + u v + v - (ds/dt)2 - Eü2 + 2FüiJ + GiJ2 .

6.17 By Exercises 5.4 and 6.3, we have (in an obvious notation),:F] = JtT]J,:Fu = ±JtTuJ, where the sign is that of det(J) . The principal curvaturesof Ö' are the roots of det(:Fu -k:F]) =0, i.e. det(±JtTuJ-kJtTuJ) =0,which (since J is invertible) are the same as the roots of det(±Tu-kT]) =O. Hence, the principal curvatures of Ö' are ± those of 0'.Let ~Ö'ii. + ijÖ'ü be a principal vector for Ö' corresponding to the principalcurvature k. Then,

so if J (~) = (~), then ~O'u +TJO'v is a principal vector for 0' correspond

. he nri al B t ' c 8u i 8u - 8vi+ 8v ing to t e pnncip curvature K . u, smce .. = 811." + 8fJ TJ, TJ = tJii." 8ü TJ ,c i ( 8u 8v) - (8u 8v) c- - -we have ..O'u + TJO'v_ =.. 8ii.(1u + 8ii.O'v + TJ 8ü(1u + 8üO'v = ..0'11. + TJ(1ü,

which shows that eÖ'ii. + ijÖ'ü is a principal vector for 0' corresponding tothe principal curvature K.

The second part also follows from Corollary 6.2.

Solutions 301

6.18 l' = Ü(fu + iia; is a principal vector corresponding to the principal curva-

ture K ~ (FII - KF]) (:) = (~) ~ FilFII (:) = K (:) ~

(~ ~) (:) = -K ( :) ~ aü + eil = -KÜ and bü + dv = -KV. But,

N = üNu+vNv = Ü(OOu+bo'v)+V(Ct:1u+Mv) = (aü+eil)(fu+(bü+dv)(fv 'Hence, l' is principal~ N = -K(Üt:1u + V(fV) = -K'Y.From Example 6.2, the first and second fundamental forms of a surfaceof revolution are du2 + f(U)2dv2 and (jü - jiJ)du2 + fiJdv2, respectively.Since the terms dudv are absent, the vectors (fu and a v are principal; butthese are tangent to the meridians and parallels, respectively.

6.19 By Exercise 6.11, N = -Knt + TgB, so by Exercise 6.18 '1 is a line ofcurvature if and only if Tg =0 (in which case A= K n ) .

6.20 Let NI and N2 be unit normals of the two surfaces; if'Y is a unit-speedparametrisation of C, then NI = -Al'Y for some scalar Al by Exercise6.18. If Cis a line of curvature of 82, then N2 = -A2'Y for some scalar A2,and then (N l .N2)" = -Al'Y.N2 - A2'Y.Nl =0, so N l .N2 is constant along'1, showing that the angle between 81 and 82 is constant. Conversely, ifN l .N2 is constant, then N l.N2 = 0 since N l .N2 = -Al'Y.N2 = 0; thus,N2 is perpendicular to Nb and is also perpendicular to N2 as N2 is aunit vector; but l' is also perpendicular to NI and N2; hence, N2 mustbe parallel to 1', so there is a scalar A2 (say) such that N2 = -A2'Y.

6.21 (i) Differentiate the three equations in (21) with respect to w, u and v,respectively; this gives

Subtracting the second equation from the sum of the other two gives(fu.(fvw = 0, and similarly (fv .(fuw = (fw.(fuv = O.(ii) Since (fv .(fw = 0, it follows that the matrix F] for the u = Uo surfaceis diagonal (and similarly for the others) . Let N be the unit normal of theu = Uo surface; N is parallel to a v x o w by definition, and hence to (fu since(fu, (fv and (fw are perpendicular; by (i), (fvw.(fu = 0, hence (fvw.N = 0,proving that the matrix F]] for the u = Uo surface is diagonal.(iii) By part (ii), the parameter curves of each surface u = Uo are linesof curvature. But the parameter curve v = vo, say, on this surface is thecurve of intersection of the u = Uo surface with the v = Vo surface.


6.22 We have Nu = oa.; + oov, N , = ca.; + dtTv, so

F (N u.N u Nu .Nv)

rII = Nu .Nv N v·Nv

_ ( Ea2 + 2Fab + Gb2 Eac + F(ad + bc) + Gbd)- Eac + F(ad + bc) + Gbd Ee + 2Fcd + GrP

= (~ :) (; ~) (: ~) =(-FilFrdFr(-FilFII)

=FrrFi1FrFi1Fu =FrrFi1Fu-

6.23 By Example 6.2, the principal curvatures are jg - jiJ and iJ/ [ , If iJ = 0,the surface is part of a plane z = constant and no point is parabolic. Thus,iJ # 0, and every point is parabolic if and only if jg - jiJ = 0. Multiplyingthrough by iJ and using j2 + iJ2 = 1 (which implies that j j + iJg = 0), weget j = 0. Hence, f(u) = au + b, where a and bare constants. If a = 0,we have a circular cylinder; if a # 0, we have a circular cone.

6.24 On the part of the ellipsoid with z # 0, we can use the parametrisation

u(x,y) = (x,y ,z) , where z = ±rVl-~-~ . The first and second

fundamental forms are (l+ z;)dx2+2zzzydxdy+(I+z~)dy2 and (zzzdx2+

2zzydxdy+zyydy2)/ VI + zi + z~ , respectively. By Proposition 5.3(ii) , the

condition for an umbilic is that Frr = KFr for some scalar K. This leadsto the equations

Z z z = >'(1 + z; ), Zzy = >'zzZy, Zyy = >'(1 + z~) ,

where>. =KV1 + zi + z~. If x and y are both non-zero , the middle equa

tion gives >. = 1/z, and substituting into the first equation gives the contradiction p2 = r2 • Hence, either x = °or y = 0. If x = 0, the equationshave the four solutions

x = 0, y = ±qvq:

- P:, z = ±rvr:

- P: .q -r r -q

Similarly, one finds the following eight other candidates for umbilics :

~2 _ q2 ~2_q2

X = ±p 2 2' Y = 0, z = ±r 2 2 'P -r r -p

x = ±PV~ =::, y = ±qv:: =; , z = 0.

Of these 12 points, exactly 4 are real, depending on the relative sizes ofp2, q2 and r2.

Solutions

Chapter 7

303

7.1 tTu = (1,1, v), tTv = (1, -1, u), tTuu = tTvv = 0, tTuv = (0,0,1). Whenu = v = 1, we find from this that E = 3, F = 1, G = 3 and L =N = 0, M = -Ij.,fi. Hence, K = (LN - M 2)j(EG - F2) = -IjI6,H = (LG - 2MF + NG)j2(EG - F2) = IjS.,fi.

7.2 For the helicoid tT(u,v) = (vcosu,vsinu,Au), tTu = (-vsinu,vcosU,A),tTv = (cosu,sinu,O), N = (A2 + v2)-1/2(-Asinu,AcosU,-v), tTuu =(-vcosu,-vsinu,O), tTuv = (-sinu,cosu,O), tTvv = O. This givesE = A2 + v2, F = O,G = 1 and L = N = O,M = AjVA2 +v2. Hence,.K = (LN - M 2)j(EG - F2) = _A2j(A2+ V2)2.For the catenoid tT(u, v) = (cosh u cos v, cosh u sin v, u), we have a u =(sinh u cos v, sinh u sin v, 1), a; = (- cosh u sin v, cosh u cosv, 0),N = sechu(-cosv,-sinv,sinhu), tTuu = (coshucosv,coshusinv,O),tTuv=(- sinh u sin v, sinh u cos v, 0), tTvv= (- cosh u cosv, - cosh u sin v, 0).This gives E = G = cosh2 U,F =°and L = -1, M = 0, N = 1. Hence,K = (LN - M2)j(EG - F2) = -sech4u.

Alternatively, use the results of Exercise 6.15.7.3 Parametrise the surface by tT(u,v) = (u,v ,f(u,v». Then, tTu = (I,O,fu),

a; = (O,I,fv), N = (I+f~+{;)-1/2(-fu,-fv,I), tTuu = (0,0, fuu) ,tTuv = (0, o'/uv) , tTvv = (0, 0, fvv). This gives E = 1 + f~,F = fufv,G =1 + {; and L = (1 + f~ + f;)-1/2 fuu, M = (1 + f~ + J';)-1/2 fuv, N =(1+ f~ + f;)-1/2 fvv. Hence,

K = fuufvv - f~v(1 + f~ + f;)2 .

7.4 (i) From Example 7.3, K = 0 {::::::::} 6.N =°{::::::::} 6.«t + v6) x eS) = 0 {::::::::}6.(t x eS) =0. If 6 = n , 6= -lI:t + rb, t x 6 = b, so K =°{::::::::} r =°{::::::::}'Y is planar (by Proposition 2.4). If 6 = b, 6= -rn, t x 6 = -n, so againK =0 {::::::::}r =0.(ii) Let NI be a unit normal of S. Then, K =°{::::::::} NdtxNd = O. SinceNI is perpendicular to NI and NI is perpendicular to t, this conditionholds {::::::::} NI is parallel to t, i.e. {::::::::} NI = -A'Y for some scalar A. Nowuse Exercise 6.IS .

7.5 Using the parametrisation tT in Exercise 4.10, we find that E = b2 , F =O,G = (a + bCOSO)2 and L = b,M = O,N = (a + bcosO)cosO. This givesK = cosOjb{a+ bcosO), dk = (EG - F2)1/2dOdep = b{a+ bcosO)dOdep.Hence,


7.6 The first part follows from Exercises 5.3 and 6.4. The dilation multipliesE, F, G by a2 and L, M, N by a, hence H by a- 1 and K by a-2 (usingProposition 7.1).

7.7 Since (T is smooth and (Tu X a; is never zero, N = (Tu x (Tvl 11 (Tu x (Tv 11

is smooth. Hence, E, F, G, L , M and N are smooth. Since EG - F2 > 0(by the remark following Proposition 5.2), the formulas in Proposition7.1(i) and (ii) show that H and K are smooth. By Proposition 7.1(iii),the principal curvatures are smooth provided H 2 > K, i.e, provided thereare no umbilics.

7.8 At a point P of an asymptotic curve, the normal curvature is zero. ByCorollary 6.2, one principal curvature Kl ~ 0 and the other K2 ~ O. HenceK = KIK2 ~ O. On a ruled surface, there is an asymptotic curve, namelya straight line, passing through every point (see Exercise 6.12).

7.9 By Exercise 6.22, FIII = FIIF]1 FII. Multiplying on the left by F]I,the given equation is equivalent to

A2+2HA+KI=0,

where A = _F]1FI I = (: ~). By the remarks following Definition

6.1, the principal curvatures are the eigenvalues of -A. Hence, 2H =sum of eigenvalues of -A = -(a + d), K = product of eigenvalues of-A = ad - bc. Now use the fact stated in the question.

7.10 By Eq. (9) in Chapter 6, -y.-y = TtFITj by Eq. (10) in Chapter 6,N.-y = -N..:y (since N.-y = 0) = -Kn = - T tFIIT . Now, N.N = (üNu +iJNv).(üNu+iJNv) = (Nu .Nu)ü2+2(Nu.Nv)üiJ+(Nv.Nv)iJ2 = TtFIIIT.Hence, multiplying the equation in Exercise 7.9 on the left by T" and onthe right by T gives N .N+2HN.-Y+K-Y.-Y = O.JE.-yis an asymptotic curve,K n = 0 so N.-y = O. So, assuming that 'Y is unit-speed, we get N.N = -K.But Exercise 6.12 gives N = ±b, so N = =frn and N.N = r 2

.

7.11 The parametrisation is (T(U, v) = (f(u) cosv, f(u) sin v ,g(u)) , f(u) = e",g(u) =VI - e2u - cosh-1(e-u

) , -00 < u < O.(i) A parallel u = constant is a circle of radius f(u) = e", so has length271'eu •

(ii) From Example 7.2, E = I,F = O,G = f(U)2 , so dAq = f(u)dudv and

the area is J: 1fJ~oo eUdudv = 271' .

(iii) From Example 7.2, the principal curvatures are Kl = jg - iiJ =-i I iJ = _(e-2U- 1)-1/2, K2 = NIP = iJI f = (e- 2u - 1)1/2.

(iv) Kl < 0, K2 > O.7.12 (i) Setting ü = v,v = w = e-u , we have u = -lnv,v = ü so, in the

notation of Exercise 5.4, J = (~ -o~). Since J is invertible, (u, v) I-t

Solutions 305

?),~

(v,w) is a reparametrisation map. The first fundamental form in terms ofv, w is given by

(~ ~) = r (; ~) J

= (_Oi ~) (~ f(~)2) (~ -oi) = (*so the first fundamental form is (dv2 + dw2)/ w2.(ii) We find that the matrix

_ (8V 8V) ( v(w+1) 1(v2_(W+1)2))J- äU 8V - 2

- ~~ ~~ - _~(v2 - (w + 1)2) v(w + 1) ,

so the first fundamental form in terms of U and V is given by

Jt(~ O)J=(V2+ (W+ 1)2)2I= 1 I° ~ 4w2 (1 - U2 - V2)2 '

after some tedious algebra.In (i), u < °and -1r < V < 1r corresponds to -1r < V < 1r and w > 1, asemi-infinite rectangle in the upper half of the vw-plane.

To find the corresponding region in (ii), it is convenient to introduce thecomplex numbers z = v + iw, Z = U + iV. Then, the equations in (ii)

are equivalent to Z = ~+~, z = i(~!.ll)' The line v = 1r in the vw-planecorresponds to z + Z = 21r (the bar denoting complex conjugate), i.e,

i(~!ll) - i(~!.ll) = 21r, which simplifies to IZ - (1 - ~W = ~; so v = 1r

corresponds to the circle in the UV -plane with centre 1 - ~ and radius ~.

Similarly, v = -1r corresponds to the circle with centre 1 + ~ and radius

~. Finally, w = 1 corresponds to z - z = 2i, i.e. i(~!.~) + i&!.ll) = 2i.This simplifies to IZ - ~12 = ~; so w = 1 corresponds to the circle withcentre 1/2 and radius 1/2 in the UV-plane. The required region in theUV -plane is that bounded by these three circles:

306

vElementary Differential Geometry

~~--7U

7.13 Let -y(u) = (J(u) ,O,g(u)) and denote d/du by a dot; by Eq. (2), j +KI =0. If K < 0, the general solution is 1 = ae-FKu + beFKu where a, bare constants; the condition 1(1r/2) = I( -1r/2) = °forces a = b = 0,so -y coincides with the z-axis, contradicting the assumptions. If K = 0,1 = a + bu and again a = b = °is forced. So we must have K > °and1 = acosVKu + bsinVKu. This time, 1(1r/2) = 1(-1r/2) = °and a,bnot both zero implies that the determinant

Icos ..jK1r /2 sin ..jK1r /2 I= 0.cos VK1r/2 - sin VK1r/2

This gives sin VK1r = 0, so K = n 2 for some integer n =!' 0. If n = 2kis even, 1 = bsin 2ku, but then 1(0) = 0, contradicting the assumptions.If n = 2k + 1 is odd, 1 = acos(2k + l)u and 1(1r/2(2k + 1)) = 0, whichcontradicts the assumptions unless k =°or -1, Le. unless K = (2k+ 1)2 =1. Thus, 1 = acosu, 9 = )1- j2 = J1- a2sin2u. Now, 'Y = <i,O,g) is

perpendicular to the z-axis {::::} 9 = 0. So the assumptions give '-"I - a2 =0, i.e. a =± 1. Then, -y(u) = (± cosu, 0, ± sin u) (up to a translation alongthe z-axis) and S is the unit sphere.

7.14 Let ü(ü, v) be a patch of S containing P =ü(üo,vo). The gaussian curvature K of S is <°at P; since K is a smooth function of (ü, v) (Exercise7.7), K (u,v) < °for (u,v) in some open set Ü containing (uo,vo); thenevery point of ü(Ü) is hyperbolic. Let 11':1,11':2 be the principal curvaturesof ü, let °< () < tt /2 be such that tan () = J -1I':t!11':2, and let Cl and e2 bethe unit tangent vectors of Ü making angles () and -(), respectively, withthe principal vector corresponding to 11':1 (see Corollary 6.1). Applying

Solutions 307

Proposition 7.4 gives the result.7.15 See the proof of Proposition 9.5 for the first part.

The first fundamental form ofthe given surface patch is (l+u2+v2)2(du2+

dv2), so it is conformal, and Uuu +utItI = (-2u, 2v, 2) + (2u, -2v, -2) = O.7.16 Parametrize the surface by u(u, v) = (u, v, f(u, v)). By Exercise 7.3, E =

1 + f~,F = fuftl,G = 1 + {;,L = (1 + f~ + J';)-1/2fuu,M = (1 + f~ +{;)-1/2 futl, N = (1+ f~ + {;)-1/2 ftltl. Hence,

H = LG - 2MF + NE = fuu(1 + {;) - 2futlfuftl + ftltl(1 + f~)

2(EG - F2) 2(1 + f~ + f';)3/2

Taking f(u,v) = In (~~:~) gives

H _ sec2 u(1 + tan2 v) - sec2 v(1 + tan2 u) _ 0- 2(1 + tan2 u + tan2 v)3/2 - .

7.17 Eu = Uu + wNu, E tl = a; + wNtI, E w = N . Eu.Ew = 0 sinceuu .N = Nu.N = 0, and similarly EtI .Ew = O. Finally, Eu .EtI =Uu.UtI +w(uu .NtI + utI.Nu) + w2Nu.NtI = F - 2wM + w2Nu.NtI =w2Nu.NtI; byProposition 6.4, Nu = -~uu, N, = -~UtI, so Nu .NtI = ~~F = O.Every surface u = Uo (a constant) is ruled as it is the union of the straightlines given by v = constant; by Exercise 7.4(ii) , this surface is Hat providedthe curve 'Y(v) = u(Uo, v) is a line of curvature of S, i.e, if U tI is a principal vector; but this is true since the matrices ;:1 and ;:11 are diagonal.Similarly for the surfaces v = constant.

7.18 By Eq. (15), the area of u(R) is

J111 Nu x x, 11 dudv =J11K'1I UU x Utl 11 dudv =J11K1dAq.

7.18 From the formula for K in the solution of Exercise 7.5, it follows that S+and S- are the annular regions on the torus given by -71"/2 ~ u ~ 71"/2and 71"/2 ~ u ~ 371"/2, respectively.

S+ S-


It is clear that as a point P moves over S+ (resp. S-), the unit normal at Pcovers the whole of the unit sphere. Hence, 11s+ IKldA = 11s - IKldA =411" by Exercise 7.18; since IKI = ±K on S±, this gives the result .

Chapter 8

8.1 By Exercise 4.4, there are two straight lines on the hyperboloid passingthrough (1,0,0); by Proposition 8.3, they are geodesics. The circle givenby z = 0, x2 + y2 = 1 and the hyperbola given by y = 0, x2 - z2 = 1are both normal sections, hence geodesics by Proposition 8.4 (see alsoProposition 8.5).

8.2 Let IIs be the plane through "Y(s) perpendicular to t(s); the parametercurve s = constant is the intersection of the surface with IIs . From thesolution to Exercise 5.17, the standard unit normal of t1' is N = -(cosOn+sin 0 b). Since this is perpendicular to t, the circles in question are normalsections.

8.3 Take the ellipsoid to be ~ + ~ + ~ = 1; the vector ('ir,~,~) is normalto the ellipsoid by Exercise 4.16. If "Y(t) = (f(t),g(t),h(t» is a curve on

j2 · 2 . 2 ,2 2 2 1the ellipsoid, R =(~+ ~ + ~ )-1 /2, 8 = (j;4 + ?- + ~ )-1 2. Now, "Y is a

geodesie -<==>;Y is parallel to the normal -<==> (1,9, h) = >..({;t ,~ , ~) for

some scalar >..(t). From ~ +;; + ~ = 1 we get fJ + ~ + I;4 = 0, hencej2 · 2 ;'2 I/ .. hh . j2 . 2 ;'2 (/2 2 h2)~+~+~+~+'+j:"2""=0, i.e. ~+~+~+>.. j;4 + ~ + j:4 =0,which gives >.. = _82 / R 2 • The curvature of"Y is

1I;Y1I=(P+92+ h2)1 /2=1>"I(P +g2 +h2)1/2 =~=!.-.p4 q4 r4 8 R2

Finally,

1d ( 1) (li 99 hh) (i29

2h2)

2dt R282 = p4 + q4 + -;:4 p2 + q2 + r2

+ (P + g2 + h2) (ii + 99 + hh)

p4 q4 r4 p2 q2 r2

1 (li 99 hh) x (li 99 hh)= R2 p4 + q4 + -;:4 + 82 p4 + q4 + -;:4 = 0,

since >.. = _82/ R2 • Hence, R8 is constant.8.4 If "Y is a geodesic, ;Y = xn is parallel to N (in the usual notation), so

n = ±N. In the notation of Exercise 6.11, B = t x N = ±b, so B =

Solutions 309

Kgt - TgN = ±b = =fTn = -TN. Hence, Tg = T (and K g = 0, whieh weknew already).

8.5 ;y is parallel to II since "I lies in II , and ;y is parallel to N since "I is ageodesic ; so N is parallel to II . It follows that N is also parallel to II .Since N is perpendieular to N, and l' is also perpendieular to N andparallel to II, N is parallel to 1'. By Exercise 6.18, "I is a line of curvatureof S.

8.6 Ir P and Q lie on the same parallel of the cylinder, there are exactlytwo geodesies joining them, namely the two circular arcs of the parallelof whieh P and Q are the endpoints. Ir P and Q are not on the sameparallel, there are infinitely many circular helices joining P and Q (seeExample 8.7).

8.7 Take the cone to be O'(u,v) = (ucosv,usinv,u) . By Exercise 5.5, 0' isisometrie to part ofthe xy-plane by O'(u,v) t-+ (uV2cos~, uV2sin .12,0).By Corollary 8.2, the geodesies on the cone correspond to the straight linesin the xy-plane. Any such line, other than the axes x =°and y = 0, hasequation ax +by = 1, where a, b are constants; this line corresponds to the

curve v t-+ ( cos v sinVI).y'2(acos 72+bsin 72)' y'2(acos ~+bsin~)' V'2(acos ~+bsin~) ,

the x and y-axes correspond to straight lines on the cone.8.8 From Example 5.3, O'(u, v) ='Y(u) + va, where "I is unit-speed, 11 a 11 =1,

and "I is contained in aplane perpendicular to a ; the map 0'(u, v) t-+

(u,v ,O) is an isometry from the cylinder to the xy-plane. A curve t t-+

O'(u(t), v(t)) is a geodesie on the cylinder {::::::} t t-+ (u(t) ,v(t) ,0) is aconstant-speed parametrisation of a straight line in the plane {::::::} üand iJ are constant {::::::} iJ is constant (since ü2 + iJ2 is constant). Sincea.ftO'(u(t) ,v(t)) = ii, iJ = constant {::::::} the tangent vector ftO'(u ,v)makes a fixed angle with the unit vector a parallel to the axis of thecylinder.

8.9 Take the cylinder to be O'(u, v) = (cosu, sin u, v) . Then, E = G = 1, F = 0,so the geodesie equations are ü = ii = 0. Hence, u = a + bt, v = c + dt,where a, b,c, d are constants. Ir b =°this is a straight line on the cylinder ;otherwise, it is a circular helix.

8.10 E = 1, F = 0, G = 1 + u2, so "I is unit-speed {::::::} ü2 + (1 + U2)iJ2 = l.

The second equation in (2) gives ft«1 + u2 )iJ ) =0, i.e. iJ = 1':u2' where2

a is a constant. So ü2 = 1 - (1~U2) and , along the geodesic,

dv = '!!. = ± adu ü J(1 - a2 +u2)(1+u2) •

Ir a = 0, then v = constant and we have a ruling. If a = 1, then dv/du =±1/uVI+ u2 , which can be integrated to give v = Vo =f sinh"? ~, where


Vo is a constant.8.11 We have

N x a = (tTu X tTtI) x tTu = (tTu.tTu)tTtI - (tTu.tTtI)tTu = EtTtI - FtTuu 11 a u X tT tI 11 J EG - F2 J EG - F2 '

and similarly for N x tTtI. Now, -y = iur; + iJtTtI, so

N. u(EtTtI - FtTu) + iJ(FtTtI - GtTu)

X 'Y - --'-----.:..---:~===1=~=---~- JEG-F2 '

;y = ÜtTu + iitTtI + U2tTu u + 2uiJtTutl + iJ2 tTtl tI .

Hence, /\,g =;Y.(Nx-Y) = (uii-iJü)VEG - F2+Au3+Bu2iJ+CuiJ2+DiJ3,

where A = tTuu .(EtTtI - FtTu) = E«tTu.tTtI)u - tTu.tTUtl) - ~F(tTu.tTu)u =E(Fu - ~EtI) - ~FEu, with similar expressions for B,C,D.

8.12 We have

(Eu2+2FuiJ+ GiJ2)'

=(Euu + EtliJ) u2 + 2(Fuu + FtliJ)uiJ + (Guu + GtliJ)iJ2

+ 2Euü + 2F(uii + üiJ) + 2GiJii

=u(Euu2 + 2FuuiJ + GuiJ2) + iJ(Etlu2+ 2FtluiJ + GtliJ2)

+ 2Euü + 2F(uii + üiJ) + 2GiJii

=2(Euu + FiJ)'u+ 2(Fu + GiJ)'iJ

+ 2(Eu + FiJ)ü + 2(Fu + GiJ)ii (by the geodesie equations)

=2[(Eu + FiJ)u]' + 2[(Fu + GiJ)iJ]'

=2(Eu2 + 2FuiJ+ GiJ2rHence, (Eu2 + 2FuiJ+ GiJ2)' = 0 and so 11 -y 11 2 = Eu2 + 2FuiJ + GiJ2 isconstant.

8.13 They are normal sections.8.14 Every parallel is a geodesie {:::::::} every value of u is a stationary point

of f(u) (in the notation of Proposition 8.5) {:::::::} f = constant {:::::::} thesurface is a circular cylinder.

8.15 The two solutions of Eq. (13) are v = Vo ± ) b - w2 , so the condition

for a self-intersection is that, for some w > 1,2)b - w2 = 2k1r for some

integer k > O. This holds {:::::::} 2)b - 1 > 21r, i.e. n < (1 + 1r2)- 1/2. Inthis case, there are k self-intersections, where k is the largest integer such

that 2k1r < 2)b - 1.

8.16 (i) If 'Y(t) is a geodesic, so is fh(t)) , and if'Yis defined for all -00 < t < 00 ,

so is f('Y(t)). So f takes meridians to meridians, i.e, if v is constant, so isv. Hence, v does not depend on w.

Solutions 311

(ii) f preserves angles and takes meridians to meridians, so must takeparallele to parallels. Hence, tÜ does not depend on v .(iii) The parallel w = constant has length 27r/w by Exercise 7.11(i) (w =e- U

) . As f preserves lengths, part (ii) implies that 27r[ui = 27r[ib, sow=w.(iv) We now know that f(O'(u,v)) = O'(F(v),w) for some smooth function

F(v) . The firstfundamental form ofO'(F(v) ,w) is w-2 ((~~)2 dv2 + dw2 ) ;

since f is an isometry, this is equal to w- 2(dv2+dw2 ) , hence dF/dv = ±1,so f(v) = ±v+ o, where o is a constant. If the sign is +, fis rotation byo around the z-axis ; if the sign is -, f is reflection in the plane containingthe z-axis making an angle a/2 with the xz-plane.

8.17 From the solution to Exercise 7.12, Z = U + iV = ~+~, where z = v +iw. Since the geodesics on the pseudosphere correspond to straight linesand circles in the vw-plane which are perpendicular to the v-axis, theycorrespond in the UV-plane to straight lines and circles perpendicular tothe image of the v-axis under the transformation z ~ ~+~' i.e, the unitcircle U2 + V2 = 1.

8.18 (i) Let the spheroid be obtained by rotating the ellipse ~+~ = 1 aroundthe z-axis, where a, b > O. Then, a is the maximum distance of a point ofthe spheroid from the z-axis, so the angular momentum {] of a geodesiemust be ::; a (we can assurne that {] ~ 0). If {] = 0, the geodesie is ameridian. If 0 < {] < a, the geodesie is confined to the annular region

on the spheroid contained between the circles z = ±bJ1 - ~' and thediscussion in Example 8.9 shows that the geodesie 'bounces' between thesetwo circles:


If n = a, Eq. (10) shows that the geodesie must be the parallel z = O.(ii) Let the torus be as in Exercise 4.10. If n = 0, the geodesie is a meridian(a circle). If 0 < n < a - b, the geodesie spirals around the torus:

If n = a - b, the geodesie is either the parallel of radius a - b or spirals around the torus approaching this parallel asymptotically (but nevercrossing it) :

If a - b < n < a + b, the geodesie is confined to the annular regionconsisting of the part of the torus a distance ~ n from the axis, andbounces between the two parallels which bound this region:

If n =a + b, the geodesie must be the parallel of radius a + b.

Solutions 313

8.19 From Exercise 5.5, the cone is isometrie to the 'sector' S of the plane withvertex at the origin and angle nV2:

Geodesies on the cone correspond to possibly broken line segments in S: ifa line segment meets the boundary of S at a point A, say, it may continuefrom the point B on the other boundary line at the same distance as Afrom the origin and with the indieated angles being equal :

o B

(i) TRUE: if two points P and Q can be joined by a line segment in Sthere is no problem; otherwise, P and Q can be joined by a broken linesegment satisfying the conditions above:

p

o s


To see that this is always possible , let PI, P2, ql and q2 be the indicateddistances, and let R and S be the points on the boundary of the sector ata distance (1J2ql +PIq2)/(1J2 + q2) from the origin. Then, the broken linesegment PR followed by SQ is the desired geodesic.(ii) FALSE:

Q

p

(iii) FALSE: many meet in two points, such as the two geodesics joiningP and Q in the diagram in (ii).(iv) TRUE: the meridians do not intersect (remember that the vertex ofthe cone has been removed), and parallel straight lines that are entirelycontained in S do not intersect.(v) TRUE: since (broken) line segments in S can c1early be continuedindefinitely in both directions.(vi) TRUE: a situation of the form

in which the indicated angles are equal is c1early impossible. But theanswer to this part of the question depends on the angle of the cone: ifthe angle is 0:, instead of 1r/4, lines can self-intersect if 0: < 1r/6, for thenthe corresponding sector in the plane has angle < 1r:

Solutions 315

o8.20 (i) This is obvious if n ~ 0 since e-1/

t 2 ~ 0 as t ~ O. We prove thatt - n e- 1/

t 2 ~ 0 as t ~ 0 by induction on n ~ O. We know the result ifn = 0, and if n > 0 we can apply L'Hopital's rule :

. r» . nt-n-1 • n t-(n-2)hm l/t2 = hm 2 l/t2 = hm -2 l/t2'e-su e t-tO i!"e t-tO e

which vanishes by the induction hypothesis.(ii) We prove by induction on n that 0 is n-times differentiable with

~O = { ~n3lt) e-1/

t 2 if t i:- 0,dtn 0 if t = 0,

where Pn is a polynomial in t . For n =0, the assertion holds with Po =1.Assuming the result for some n ~ 0,

~+l0 _ (-3nPn P~ 2Pn ) e-1/t2dtn+1 - t3n+l + t3n + t3n+3

if t i:- 0, so we take Pn+l =(2 - 3nt2)Pn + t3P~ . If t =0,

~+10 _ . Pn(t) -1/t2 _ . e-1/t2_

d +1 - hm 3 + 1 e - Pn(O) hm 3 +1 - 0tn t-tO t n t-tO t n

by part (i).Parts (iii) and (iv) are obvious .

8.21 Since "(s is unit-speed, Ur'Ur = 1, so JoR

Ur 'U r dr = R. Differentiating

with respect to 0 gives JoR

Ur.UrS dr = 0, and then integrating by partsgives

lR

r=Rus.urlr=o - 0 US 'U rr dr = O.

Now u(O,O) = P for all 0, so Us = 0 when r = O. So we must show thatthe integral in the last equation vanishes. But, U r r = 'Ys, the dot denotingthe derivative with respect to the parameter r of the geodesie "(s, so Urr

is parallel to the unit normal N of Uj since us.N = 0, it follows thatUS'Urr = O.The first fundamental form is as indicated since Ur 'U r = 1 and Ur .Us = O.

316

Chapter 9


9.1 This follows from Exerdse 7.6.9.2 This follows from Exercise 7.3.9.3 K1 + K2 = 0 and K1 = K2 ==> K1 = K2 = O.9.4 K1 + K2 = 0 ==> K2 = -K1 ==> K = K1K2 = -K~ :$ O. K = 0~ K~ =

o~ K1 = K2 = 0~ the surface is part of a plane (by Proposition6.5).

9.5 By Proposition 7.6, a compact minimal surface would have K > 0 at somepoint, contradicting Exerdse 9.4.

9.6 By the solution of Exerdse 7.2, the helicoid 0'(u, v) = (v cos u, v sin u, >.u)has E = >.2 + v2 , F = O,G = I ,L = O,M = >./(>.2 + V 2)1/2,N = 0, so

H = LG- 2MF+NG =02(EG - F2) .

9.7 A straightforward calculation shows that the first and second fundamentalforms of O't are cosh'' u(du2 +dv2 ) and - cos t du2 - 2 sin t dudv + cos t dv2 ,

respectively, so

H = - cos t cosh2

U + cos t cosh2

U = O.2cosh4 u

9.8 From Example 4.10, the cylinder can be parametrised by O'(u, v) = 'Y(u) +va , where 'Y is unit-speed, 11 a 11 = 1 and 'Y is contained in aplane IIperpendicular to a. We have O'u = 'Y = t (a dot denoting d/du), a; = a,so E = 1, F = 0, G = 1; N = t x a , O'uu = i = xn, O'uv = O'vv = 0, soL = Kn.(t x a) , M = N = O. Now t x a is a unit vector parallel to IIand perpendicular to t, hence parallel to n; so L = ±K and H = ±K/2.SO H = 0~ K = 0~ 'Y is part of a straight line~ the cylinder ispart of a plane.

9.9 Using Exercise 9.2, the surface is minimal ~

(1 + g'2)j + (1 + p)g" = 0,

where a dot denotes d/dx and a dash denotes d]dy ; hence the statedequation. Since the left-hand side of this equation depends only on x andthe right-hand side only on y , we must have

j g"--. =a, ---=-a1 + j2 1 + g/2 '

for some constant a. Suppose that a =j:. O. Let r = j ;then j =rdr/df andthe first equation is rdr / df = a(1 + r2 ) , which can be integrated to giveaf = ! In(1 + r 2 ) , up to adding an arbitary constant (which correspondsto translating the surface parallel to the z-axis). So df /dx = ±Je2aJ - 1,

Solutions 317

which integrates to give f = -~ In cosa(x + b), where b is a constant;we can assurne that b = 0 by translating the surface parallel to the xaxis , Similarly, 9 = ~ In cos ay, after translating the surface parallel to they-axis. So, up to a translation, we have

_ 1 1 (cosay )z--n -- ,a cosax

which is obtained from Scherk 's surface by the dilation (x, y, z) H

a(x, y, z) . Ir a = 0, then j = 9" = 0 so f = b + cx, 9 = d + ey, forsome constants b, c, d, e, and we have the plane z = b+ d + cx + ey.

9.10 The first fundamental form is (cosh v + 1)(cosh v - cos u)(du 2 + dv2 ) , so ais conformal. By Exercise 7.15, to show that a is minimal we must showthat lTuu + lTvv = 0; but this is so, since

lTu u = (sin u cosh v, cos u cosh v, sin ~ sinh ~),

lTvv = (- sin u cosh v, - cosu cosh v, - sin ~ sinh ~) .

(i) lT(O, v) = (0,1 - cosh v ,0), which is the y-axis . Any straight line is ageodesic.(ii) lT(11", v) = (11", 1+cosh v, -4 sinh ~), which is a curve in the plane x =11"such that

Z2 = 16sinh2 ~ = 8(coshv -1) = 8(y - 2),

i.e, a parabola. The geodesie equations are

d(E ') 1E ( ' 2 ' 2) d(E ') 1E ( ' 2 ' 2)dt u = 2 u u +v , dt v = 2 v u +v ,

where a dot denotes the derivative with respect to the parameter t of thegeodesie and E = (cosh v +1)(cosh v - cos u). When u =11", the unit-speedcondi tion is EiJ2 = 1, so iJ = 1/(cosh v + 1). Hence, the first geodesieequation is 0 = !Eu iJ2, which holds because Eu = sin u(cosh v + 1) = 0when u = 11"; and the second geodesie equation is

~ (cosh v + 1) = (cosh v + 1) sinh v iJ2 = sinh v b,

which obviously holds.(iii) lT(u, 0) = (u-sin u, 1-cosu, 0), which is the cycloid ofExercise 1.7 (inthe xy-plane, with a = 1 and with t replaced by u). The second geodesieequation is satisfied because E; = sinh v(2 cosh v + 1 - cos u) = 0 whenv =O. Theunit-speedconditionis2(1-cosu)ü2 = l,soü = 1/2sin~. Thefirst geodesie equation is 1t (4 sin2 ~ü) = sin uü2 , i.e, 1t (2 sin ~) = cos ~ü,which obviously holds .

9.11 >. = a2 + bc = -(ad - bc) (since d = -a) = - det W = - K.


9.12 (i) From Example 9.1, N = (-sech u cos v, -sech u sin v, tanh u). Hence ,if N(u,v) = N(U',V'), then u = u' since u ~ tanhu is injective, socos v = cosVi and sin v =sin o" , hence v = Vi; thus, N is injective. If N =(x , y , z), then x2 + y2 = sech2u :p 0, so the image of N does not containthe poles. Given a point (x, y, z) on the unit sphere with x2 + y2 :p 0, letu = ±sech-1 J x2 + y2, the sign being that of z, and let v be such thatcosv = -x/Jx2 +y2, sinv = -y/Jx2 +y2 j then, N(u,v) = (x ,y,z).(ii) By the solution ofExercise 7.2, N =(A2+V2)- 1/2(-Asin u, Acos u, -v) .Since N (u, v) = N (u + 2k7r, v) for all integers k, the infinitely many points0'(u +2k1l", v) = 0'(U, v) + (0, 0, 2k1l") of the helicoid all have the same imageunder the Gauss map. If N = (x, y, z), then x2 + y2 = A2/(A2 + v2) :p 0,so the image of N does not contain the poles. If (x, y, z) is on the unitsphere and x2 + y2 :p 0, let v = -AZ/Jx2 + y2 and let u be such thatsinu = -x/Jx2 +y2, COSU = -y/Jx2 +y2; then N(u, v) = (x,y,z).

9.13 The plane can be parametrised by O'(u,v) = ub +vc, where {a, b,c} is aright-handed orthonormal basis of R 3

. Then, tp = O'u - ia; = b - ic . Theconjugate surface corresponds to itp = c + ib; since {a, c, - b} is also aright-handed orthonormal basis of R 3

, the plane is self-conjugate (up toa translation).

9.14 tp = (~f(1 - g2), ~ f(1 + g2), fg) ==} itp = (~if(1 - g2), ~if(1 + g2), ifg),which corresponds to the pair if and g.

9.15 By Example 9.6, tp«() = (sinh (, -i cosh (,1). From the proof of Proposition 9.7, f = <PI - i<P2 =sinh( - cosh( = -e-C:, 9 = <P3/f = -eC:.

9.16 cp = O'u - ia; = (1 - u2 + v2 - 2iuv, 2uv - i(1 - v2 + u2 ) , 2u + 2iv) =(1 - (2, -i(1 + (2), 2() . So the conjugate surface is

Ö'(u, v) =~e! (i(1 - (2),1 + (2, 2i() d(

=~e (i (( - ~) , (+ ~ ,i(2) (up to a translation)

= (-v + u2v _ v3

U+ u3

_ uv 2 -2UV)3' 3 ' .

Let U = (u - v)/V2, V = (u + v)/V2, Ö'(U, V) =O'(u,v); then,

- ( 1 ( 2 2 1 3 1 3)O'(U V) = - U - V + UV - U V + - V --U, V2 3 3 '

~ (u + V + UV2 + U2V - !V3 - !U3) U2 - V2) .V2 3 3 '

Applying the 11"/4 rotation (x,y,z)~(~(x + y), ~(y - x),z) to Ö'(U, V)

then gives (U - ~U3 + UV2, V - ~V3 + U2V, U2 - V 2), which is Enneper's surface again.

Solutions 319

9.17 I{) = (t(1- C 4)(1 - (2) , t(1 - (-4)(1 + (2) , «(1 - (-4)), SO

(1 ( (3 -1 (-3) i ( (3 -1 (-3) (2 (-2)

u=~t 2 (-3"-( +3 '2 (+3"+( +3 '2+2

= ~t ( -~( - C 1) 3 , ~( + C 1)3, ~( + C 1)2),

up to a translation. Put ( = i, ( = ü+iii. Then, u(u, v) = ü(ü, ii), where

ü(ü , ii) = ~t ( -~ sinh3( , ~ cosh" ( ,2 cosh'' ()

= (4sinhüCOSii(COSh2üsin2ii - ~sinh2ücos2ii),

4 sinh ü sin ii(~ sinh2ü sin2ii - cosh2Ü cos2ii),

2(cosh2Ü cos2ii - sinh2ü sin2ii)) .9.18 By Example 5.7, 1r(x,y,z) = (u,v,O), where x = 2u/(u2 +v2 + l),y =

2v/(u2 + v2 + 1), z = (u2 + v2 - 1)/(u2 + v2 + 1). If ( = u + iv , thenx+iy = 2(/(1(12+1), z = (\(12_ 1)/ (\( 12+ 1). This gives ( = (x+iy)/(I-z).Hence, identifying (u, v ,0) with u+iv, we have 1r(x, y, z) = (x+iy)/(I-z).From Eq. (25) , Q() = IgIJ+l (g + g,-i(g - g), Igl2 - 1), so

g+g+g-g1r(Q()) = Igl2 + 1 _ (lgl2 - 1) =g(() .

Chapter 10

10.1

10.2

By Corollary 10.2(i),

1 (0 ((e>.)u) 0((e>')tI)) 1K=-2e>' ou ~ + ov ~ =-2e>.(Auu+AtltI).

. (E F) (E F) (8r 8r)By Exercise 5.4, F G = Jt F G J, where J = !~ !~ =

( !tI ~). By Exercise 8.21 , E = 1, F = 0, and we get the stated?"" ~ --- - 2a - » ra

formulas fo: E ,F,G. Fr~m E - 1 = 2' (~-J), G - 1 = ~ (~-1),

we get u2 (E - 1) = v2 (G - 1). Since E and G are smooth functions of(u, v), they have Taylor expansions E = L:i+ ' < 2 eijUiVj + o(r2), G =L:i+j<2 gijUiVj +o(r2), where o(rk) denotes ter~s such that o(rk)/rk -+oas r -+ O. Equating coefficients on both sides of u2 (E -1) = v2 (G -1)


shows that all the e's and g's are zero except e02 = g20 = k , say. Then,jj; = 1 + kv2 + 0(r2), which implies that G = r 2 + kr" + 0(r4).

By Corollary 1O.2(ii), K = - Ja 8~'!P . From the first part , .j(j = r +~ kr3 + 0(r3), hence K = -3k + 0(1). Taking r = 0 gives K(P) = -3k.

10.3 (i) CR = J;7T 11 Uo 11 d8 = J;7T .j(jdB = J;7T (R - ~K(P)R3 + 0(R3)) dB= 211" (R - ~K(P)R3 + 0(R3)).

(ii) Since dAq = .j(jdrdB, the area AR = JoR J027T

.j(j drdB is equal to

211" JoR (r - ~K(P)r3 + 0(r3)) dr = 1I"R2 (1- Kg')R2 + 0(R2)).

Let S be the unit sphere, let P be the north pole, and let (B ,cp) bethe usual latitude longitude parametrisation of S. Then, the geodesiecircle with cent re P and radius R is the parallel B = ~ - R , which is anordinary circle of radius sin R, so that CR = 211" sin R; since 211" sin R =211"(R - ~R3 + 0(R3)) and K = 1, this agrees with the formula in (i).Similarly,

AR= r27T["/2 COSBdBdCP=211"(I-cOSR)=211"(~2 _~: +0(R4)),

Jo J7T/2 -R

in agreement with the formula in (ii).10.4 (i) Let s be the arc-length of 'Y, so that ds/dB = A, and denote dlds by

a dot. The first of the geodesie equations «2) in Chapter 8) applied to'Y gives r = ~GrÖ2 . Since r = !(B) , this gives

1 (1)' 1A AI' = 2A2Gr.

This simplifies to give the stated equation.(ii) Since Ur and-j are unit vectors , cos1/J = ur.'Y = ±Ur.(f'ur + uo) =

I'/A. Also, Ur X 'Y = ±(ur X uo) = VfN , so sin1/J = .j(j/A. Hence ,

( ~ ) 1 = -1/J' sin 1/J = - ~1/J/,

1/J' = __1 (I" _f'A') = __1_8G = _ 8.j(j.j(j A 2.j(j 8r 8r .

(iii) Using the formula for K in Corollary 1O.2(ii) and the expression forthe first fundamental form of U in Exercise 8.21, we get

J! KdAq = rL A t'" __1_82vpVGdrdB

A BC Jo Jo .j(j 8r

lL A 8.j(j r=f(O) l L A

(' 8.j(j I )= - -- dB = 1/J + -- dB.o 8r 0 8r

r=O r=O

Solutions

By Exercise 10.2, JG = r + o(r) so ßJG/ßr = 1 at r = O. Hence,

J'( KdAa = 1/7(LA) - 1/7(0) + LA = LC - (11" - LB) + LAlABe

=LA + LB + LC - 11" .

,p(L A )

321

,p(O)

B

10.5 By Exercise 5.7, generalised cylinders and cones are isometrie to a plane .Hence, if the sphere were isometrie to a generalised cylinder or cone, thesphere would be isometrie to a plane, sinee eomposites of isometries areisometries. This eontradiets Proposition 10.l.

10.6 With the notation of Example 4.9 we have, on the median circle t = 0,a t = (- sin ~ eos (}, - sin ~ sin (}, eos ~) , oU = (- sin (}, eos (}, 0), henee E =1,F = O,G = 1 and N = (-eos~eos(},-sin~sin(},-sin~); qtt = 0,a tU = (- ~ eos ~ cos (} + sin ~ sin (}, - ~ eos ~ sin (} - sin ~ eos (}, - ~ sin ~) ,

from whieh L = 0, M = ~. Henee, K = (LN - M2)/(EG - F 2) = -1/4.Since K :f 0, the Theorema Egregium implies that the Möbius band isnot isometrie to a plane.

10.7 The first fundamental forms of a and iT are found to be (1+~ )du2+u2dv2

and du2 + (u2 + 1)dv2, respectively. Since these are different, the mapq(u,v) I-t iT(u, v) is not an isometry. Nevertheless, by Corollary 1O.2(ii),the gaussian curvatures are equal:

K=K= 2~(:U(~))=-(1+lU2)2.If q( u, v) I-t iT(ü, v) is an isometry, the Theorema Egregium tells usthat -1/(1 + U2)2 = -1/(1 + Ü2)2, so Ü = ±u; let v = f(u, v). Thefirst fundamental form of iT(±u, f(u, v)) is (1 + (1 + u2)f~)du2 + 2(1 +u2)fufv dudv+(1+u 2)f;dv2j this is equal to the first fundamental form ofq(u,v) <==> 1+(1+u2)f~ = 1+ 1/u2, fufv =°and (1+u2)f; = u2. Themiddle equation gives fu =°or fv = 0, but these are both impossibleby the other two equations. Henee, the isometry does not exist.

10.8 For the eatenoid q(u,v) = (eoshueosv,eoshusinv,u), the first fundamental form is cosh'' u(du2+dv2) (Exercise 5.8) and the gaussian curvature is K = -seeh4u (Exercise 7.2). If q(u, v) I-t q(ü,v) is an isometry,


then sech4u = sech4ü, so Ü = ±Uj reflecting in the plane z = 0 changesU to -U, so assume that the sign is +. Let ii = f(u , v); the first fundamental form of O'(±u, f(u , v)) is (cosh'' U + f~)du2 + 2fufv cosh2 ududv +f; cosh" udv2

j hence, cosh'' u = cosh'' U + f~ , fufv = 0 and f; cosh2 u =cosh2

U , so I« = 0, fv = ±1. Thus, f = ±v+ Q , where Q is a constant; ifthe sign is +, we have a rotation by Q about the z-axis: if the sign is we have arefleetion in the plane containing the z-axis, making an angleQ/2 wit h the xz-plane.

-1 (cos2V 0) -1 (- cos?V 0)

10.9 W = -Tl Tu = - 0 1 0 -1 = I , so Nu = O'u,

N, = O'v' Thus, N = O'-a, where a is a constant vector. Hence, 11 O'-a 11

= 1, showing that the surface is part of the sphere of radius 1 and cent rea . The standard latitude longitude parametrisation of the unit sphere hasfirst and second fundamental forms both given by du2+ cos2 vdv 2, so theparametrisation O'(v , u) has the given first and second fundamental forms(the second fundamental form changes sign because 0' v X 0' u = -0'u X 0' v).

10.10 r i2 = sin u cos u and the other Christoffel symbols are zero; the secondCodazzi-Mainardi equation is not satisfied.

'10.11 The Christoffel symbols are rl1 = Eu/2E , rrl = -Ev/2G, rl2 =

Ev/2E, rr2 = Gu/2G, Fi2 = -Gu/2E, ri2 =Gv/2G. The first CodazziMainardi equation is

LEv (-Ev ) 1 (L N)L; = 2E - N 2G = 2" E; E + G '

and similarly for the other equation. Finally,

s ; (L N) LEv e, (N L) e ; ((~dv = 2E E + G - E2 = 2E G - E = 2E ~2 - ~d ,

and similarly for (~2)u'

10.12 Arguing as in the proof of Theorem 10.4, we suppose that J attainsits maximum value > 0 at some point P of S contained in a patch0' of S . We can assume that the principal curvatures ~1 and ~2 of 0'

satisfy ~1 > ~2 > 0 everywhere. Since H = H~l + ~2), ~1 > HandJ = 4(~1 - H)2. Thus, J increases with ~1 when ~1 > H, so ~1 musthave a maximum at P, and then ~2 = 2H - ~1 has a minimum there.By Lemma 10.2, K ::; 0 at P , contradicting the assumption that K > 0everywhere.

Solutions

Chapter 11

323

11.1 If 'Y is a simple closed geodesic , Theorem 11.1 gives JJ;nt('Y ) KdA = 271" ;since K ~ 0, this is impossible. The parallels of a cylinder are not t heimages under a surface patch a : U -t R 3 of a simple closed curve 1r inthe plane such that int(1r) is contained in U. Note that the whole cylindercan actually be covered by a single patch (see Exercise 4.2) in which Uis an annulus, but that the parallels correspond to circles going 'aroundthe hole' in the annulus.

11.2 By Proposition 2.2, "-s = dcp/ds, where cp is the angle through which afixed unit vector must be rotated anti-clockwise to bring it into coincidence with the unit tangent vector of 'Y. So

ll('Y) ll('Y) dcp

"-s ds = -dds = 271" ,o 0 s

by the Umlaufsatz.11.3 By Corollary 11.1, the interior angles 01, • . . , On of the polygon satisfy

tOi = (n - 2)71" + f' ~ KdAu.i=l Jint('Y)

Since 0 < 0 i < 271" for all i , the left-hand side is > 0; since K < 0, we have(n - 2)71" > 0 and hence n ;::: 3; and if n = 3, then JJ;nt('Y)(- K)dAu < 71"so

f'r dAu s f' r (-K)dAu < 71".Jint('Y) Jint('Y)

11.4 The parallel u = U1 is the curve 'Yl (v) = (f(ud cos v, f(U1) sin v ,g(Ul));if s is the arc-length of "Y1 ' dsfd» = /(ut}. Denote d/ds by a dot andd/du by a dash. Then, "I= (-sinv,cosv,O),;Y = - !(~tl(cosv,sin v ,O) ,

and the unit normal of the surface is N = (-g' cos V, -s' sin v , /,) . This

gives the geodesie curvature of "Y as "-g = ;Y.(N x "I) = ~(<::8 . Since

f('Y1) = 271"f(ud , Jol("Yl) "-gds = 271"/'(Ul) . Similarly for 'Y2' By Example7.2, K = -1"//, so

f'r r r rJR KdAu = Jo JUl -jfdudv =21l'(f'(ud - f'(U2))'

Hence ,

r: rl('Y2) f' r

J0 "-g ds - J0 "-g ds = JR K dAu.

This equation is the result of applying Theorem 11.2 to the followingcurvilinear polygon,


where AB is part of the meridian v = 0 (or v = 211"); the integrals ofK g along AB and along BA cancel out . (Strictly speaking, this curveis not a curvilinear polygon in the sense of Definition 11.2 - condition(i) is violated - but this difficulty can be circumvented by replacing thedouble path AB and BA by two meridians v = e and v = 211" - € andthen letting e tend to zero.)

11.5 3F = 2E because each face has 3 edges and each edge is an edge of 2faces. From X = V -E+F, we get X = V -E+~E, so E = 3(V -X) . Sinceeach edge has 2 vertices and two edges cannot intersect in more than onevertex, E S; ! V(V -l)j hence, 3(V - X) S; ! V(V -1), which is equivalentto V 2 -7V+6X ~ O. The roots of the quadratic are! (7± J49 - 7X), soV S; ! (7- J49 - 7X) or V ~ t (7 + J49 - 7X) . Since X = 2,0, - 2, .. .,the first condition gives V S; 3, which would allow only one triangle;hence, the second condition must hold.

11.6 The n triangles have 3n vertices , but each vertex is counted r times asit is a vertex of r triangles, so V = 3n/rj similarly, E = 3n/2. Then,V - E + F = 2 gives 6/r - 4/n = 1. This implies 6/r > 1, so r < 6.Triangulations for r = 3,4 and 5 can be obtained by 'inflating' a regulartetrahedron, octahedron and icosahedron, respectively.

tetrahedron octahedron icosahedron

Solutions 325

11.7 H such curves exist they would give a triangulation of the sphere with 5vertices and 5 x 4/2 = 10 edges , hence 2+10- 5 = 7 polygons. Since eachedge is an edge of two polygons and each polygon has at least 3 edges,3F ~ 2Ej but 3 x 7 > 2 x 10. H curves satisfying the same conditionsexist in the plane, applying the inverse of the stereographic projectionmap of Example 5.7 would give curves satisfying the conditions on thesphere, which we have shown is impossible.

11.8 Such a collection of curves would give a triangulation of the sphere withV = 6, E = 9, and hence F = 5. The total number of edges of all thepolygons in the triangulation is 2E = 18. Since exactly 3 edges meet ateach vertex, going around each polygon once counts each edge 3 times,so there should be 18/3 = 6 polygons, not 5.

11.9 By Corollary 11.3, ffs KdA = 411"(1 - g), and by Theorem 11.6, 9 = 1since S is diffeomorphic to Tl. By Proposition 7.6, K > 0 at some pointof S.

11.10 The ellipsoid is diffeomorphic to the unit sphere by the map (x ,y,x) t-+

(x / a, y / a, z / b), so the genus of the ellipsoid is zero . Hence, Corollary 11.3gives ffs KdA = 411"(1 - 0) = 411".Parametrising the ellipsoid by 0'(0, cp) = (a cos 0 cos ip, a cos 0 sin cp, bsin 0)(cf. the latitude longitude parametrisation of the sphere), the first andsecond fundamental forms of 0' are (a2sin20+b2cos20)d02+a 2cos2 Odcp2

and va 2 sin2~~b2 cos2 1J (d02 + cos20dcp2), respectively. This gives K =b2/(a2sin20+b2cos20)2, dAcr =acosOVa2sin2 0 + b2cos- OdOdcp; hence ,

/1 12"' 1 "./ 2 ab2 cosO dOdcpKdAcr = .

s 0 _". /2 (a2sin2 0 + b2cos- 0)3/2

11.11 K > 0 =? ffs KdA > O. By Corollary 11.3, 9 < 1; since 9 is a nonnegative integer, 9 = 0 so S is diffeomorphic to a sphere by Theorem11.6. The converse is false: for example, a 'cigar tube' is diffeomorphicto a sphere but K = 0 on the cylindrical part.


11.12 Both surfaces are closed subsets of R 3, as they are of the form 1-1 (0),

where I : R 3 ~ R is a continuous function (equal to x2 - y2 + Z4 - 1and x 2 + y2 + z4 - 1 in the two cases). The surface in (i) is not bounded,and hence not compact, since it contains the point (1, a2 , a) for all realnumbers a; that in (ii) is bounded, and hence compact, since X 2+y2+z4 =l~-l~x,y,z~1.

The surface in (ii) is obtained by rotating the curve x2 + z4 = 1 in thexz-plane around the z-axis:

It is clearly diffeomorphic to the sphere, so X = 2.11.13 Take the reference tangent vector field to be ( = (1,0), and take the

simple closed curve 'Y(s) = (coss,sins) . At 'Y(s), we have V = (Ci,ß),where

'ß - {(coss + i sin s)k if k > 0,Ci + t - ( " ) k if k 0cos s - t sm s - 1 <.

By de Moivre's theorem, Ci = cos ks, ß = sin ks in both cases . Hence,the angle t/J between V and ( is equal to ks, and Definition 11.6 showsthat the multiplicity is k.

11.14 If O'(u,v) = ü(ü,v), where (ü,v) M (u ,v) is a reparametrisation map,h V ß - - ß- - - 8ü ß8iJ ß- 8ü ß8iJt en =CiO'u7 O'v=CiO'ü+ O'iJ~Ci=Ciäü+ 8u' =Ci 8v+ 8v'

Hence, Ci and ß are smooth if Ci and ß are smooth.Since the components of the vectors 0' u and 0' v are smooth, if V is smoothso are its components. If the components of V = CiO'v+ßav are smooth,then V .O'u and V.O'v are smooth functions, hence

G(V.O'u) - F(V.O'v)Ci = EG -F2 '

ß = E(V.O'v) - F(V.O'u)EG-F2

are smooth functions, so V is smooth.11.15 If;P is the angle between V and e, we have ;p - t/J = () (up to multiples

of 271"); so we must show that J;m iJ ds = 0 (a dot denotes d/ds) . This isnot obvious since () is not a weIl defined smooth function of s (althoughd()/ds is weIl defined) . However, p = cos() is weIl defined and smooth,since p = (.e; 11 ( 1111 e11 · Now, jJ = -Osin(), so we must prove that

Solutions

J:('Y) -J--ds = O. Using Green's theorem, this integral is equal toy1-p 2

327

rPudu + Pvdv r 8 ( Pv ) 8 ( Pu )J'Ir VI - p2 = Jint('Ir) 8u VI - p2 - 8v VI _p2 '

where er is the curve in U such that 'Y(s) = O'('Ir(s))j and this Une integralvanishes because

8 ( Pv ) 8 ( Pu )8u ~ = 8v VI _p2

11.16 Let F : S ~ R be a smooth function on a surface S, let P be a point of S,let 0' and iT be patches of S containing P, say 0'(UO, vo) = iT(üo, vo) = P,and let I = F 0 0' and i = F 0 iT. Then, iü ~ lu$~ + Iv~, iv =lu~~ + Iv~, so if I« = Iv = 0 at (uo, vo), then l« = Iv = 0 at (üo, vo).Since I« = Iv = 0 at P, we have

s; = t.; (::r + 2/ uv:: :~ + Ivv (:~r,with similar expressions for iüv and ivv . This gives, in an obvious no

tation, il = J t1iJ, where J = (i~ i!) is the jacobian matrix of8ü ~

the reparametrisation map (ü,v) t-7 (u,v) . Since J is invertible, il isinvertible if 'H is invertible.Since the matrix 1i is real and symmetrie, it has eigenvectors VI, V2,

with eigenvalues Al , A2, say, such that vfvj = 1 if i = j and 0 if i :j; j .Then, if v = 0:1 VI + 0:2V2 is any vector, where 0:1,0:2 are scalars, vt1iv =A10:~ + A20:~ j hence, vt1iv > 0 (resp. < 0) for all v :j; 0 {::::} Al and A2

are both > 0 (resp . both < 0) {::::} P is a local minimum (resp. localmaximum); and hence P is a saddle point {::::} vt1iv can be both > 0and < 0, depending on the choiee of v. Since J is invertible, a vectorv :j; 0 {::::} v = Jv :j; O; and vtilv = vtJ t1iJv = vt1iv. The assertions inthe last sentence of the exercise follow from this.

11.17 (i) Ix = 2x - 2y, Iy = - 2x + 8y , so Ix = Iy = 0 at the origin . lxx =2,lxy = -2,/yy = 8, so 1i = (~2 ~2). 1i is invertible so the origin

is non-degenerate; and the eigenvalues 5 ± v'l3 of 1i are both > 0, so itis a local minimum.

(ii) Ix = Iy = 0 and 1i = (~ ~) at the origin; det1i = -16 < 0, so

the eigenvalues of 1i are of opposite sign and the origin is a saddle point.(iii) Ix = Iy= 0 and 1i = 0 at the origin, whieh is therefore adegeneratecritical point.


11.18 Using the parametrisation u in Exercise 4.10 (with a = 2, b = 1) givesJ(O, ip) = F(u(O, ip)) = (2 + eos 0) eos ip + 3. Then, 10 = - sin 0 eos ip,

J<P = -(2 + eosO) sin o; sinee 2 + eosO > 0, J<P = °~ ip = °orn, and then Je = °~ 0 = °or 'Trj so there are four eritical points,P = (3,0,0), Q = (1,0,0), R = (-1,0,0) and S = (-3,0,0). Next, 'H =

(- eos 0 eos ip sin 0 sin ip ) = (-1 0) at P = (1 0)

sinOsinip -(2 + eosO)eosip ° -3 ' 0-1

(-1 0) (1 0) .at Q, = ° 1 at R, and = ° 3 at S; henee, P IS a loeal

maximum, Q and R are saddle points, and S is a loeal minimum (all ofwhich is geometrically obvious).

angle, 106angular momentum, 185anti-holomorphic, 112, 219Archimedes's Theorem, 117, 238, 297arc-length, 7-8, 11, 97area, 50, 113-5, 202- of a geodesie triangle, 119, 237,

256-7, 320-1- signed, 168astroid, 3, 6, 281, 285asymptotic curve, 129-30, 150-1, 161,

304atlas, 60- maximal 68bifurcation, 271binormal, 36Catalan's surface, 216, 317catenary, 9, 283catenoid, 83, 106, 140, 150, 203, 216,

219, 223, 227, 240, 296, 300, 303, 318,321

Cauchy-Riemann equations, 112, 220,296

cent re of curvature, 35Christoffel symbols, 241, 322circular cone, 63, 72, 103, 105, 180,

196,313-5circular cylinder, 65, 103, 136, 179-80,

252,289circular helix , 26-7, 39, 104, 115, 287,

296cissoid, 15Clairaut's theorem, 183, 185closed set, 195

Index

Codazzi -Mainardi equations, 241, 244,322

compact, 164conformal, 107-8, 111, 190, 203, 216-7,

219-21, 296, 307connected, 18, 72continuous, 60convex, 55, 289Cornu's spiral, 33critical point, 275-8, 280, 327-8- non-degenerate, 276-8, 280curvature- constant, 31-2, 41, 44- of a curve, 24, 25- gaussian, 147, 150, 161-2, 164, 166,

168-9, 207, 219, 225-6, 229, 233, 239,245, 248, 268

- geodesie, 127, 248, 310, 323- mean, 147, 150, 161-2, 201, 307- normal, 127, 130, 137-8, 140, 299-300- principal, 132-3, 137-8, 140, 142-3,

148, 150, 162, 164, 244-4, 300, 304,306

- signed, 28-30, 34-6, 51, 165, 242, 252curve- asymptotic, 129-30, 150-1, 161, 304- level, 2, 16, 19, 96- parallel, 35, 285- parameter, 107,161,175- parametrised, 2, 16, 19, 21- plane, 40, 150, 242- smooth, 4, 74, 96- spherical, 6, 45, 287, 298curvilinear polygon, 252- positively-oriented, 252

329

330

cyeloid , 6, 35, 282-3diffeomorphic, 71diffeomorphism, 71- conformal, 107-8, 111, 121, 190- equiareal, 116, 121, 297dilation, 207, 304disc model, 155, 190double to rus, 274doubl y ruled, 66, 90, 290, 293Dupin's Theorem, 141edge , 252ellipsoid, 73, 85, 91, 145, 175, 268, 302,

308elliptic cylinder, 86elliptic paraboloid, 85, 93, 126, 129,

143, 294elliptic point, 143, 164Enneper's surface, 163, 214, 227epicyeloid, 6, 282equiareal, 116, 121, 297Euler number, 260-2, 264-5, 267, 269,

271,324Euler 's Theorem, 137evolute, 35-6, 285first fundamental form, 98, 100, 102,

108,242Four Vertex Theorem, 56Frenet-Serret equations, 41Fresnel's integrals, 33Gauss-Bonnet Theorem, 248, 253, 255,

260Gauss equations, 240Gauss map, 166, 169, 217, 228, 318-9gauss ian curvature, 147, 150, 161-2,

164, 166, 168-9, 207, 219, 225-6, 229,233, 239, 245, 248, 268

- constant, 162, 235, 244Gauss's Lemma, 200, 315general helix , 44-5generalised cone, 79, 89, 99, 105, 157-8,

239,296generalised cylinder, 78, 84, 89, 99,

105, 157-8, 168, 174, 180, 216, 239,295-6,309

genus, 258, 265, 325geodesic, 171, 175, 177-83, 187-96, 217,

308-15, 317- incomplete, 186, 196- simple elosed, 252, 323geodesie cirele , 199-200, 237, 320geodesie coordinates, 199, 235geodesie curvature, 127, 172, 181,248,

310,323


geodesie equations, 176-7, 181, 309,317, 320

geodesie patch, 199geodesie polar patch, 199geodesie torsion, 129, 141, 309gradient, 275graph, 72, 163great cirele, 174,179,299Green's Theorem, 50helicoid, 77, 106, 140, 150, 181, 210-1,

216, 219, 223, 227, 239, 292, 296, 300,303,318

Henneberg's surface, 227holomorphic, 112, 219-20, 222homeomorphism, 60Hopf's Umlaufsatz, 250-2, 270, 323hyperbolic cylinder, 86hyperbolic paraboloid, 72-3, 86, 89, 93,

143,293-4hyperbolic point, 143, 161, 164hyperboloid- of one sheet, 66, 85, 91, 175, 187,

289, 308- of two sheets, 85, 91hypo cyeloid, 6, 282integral curve, 269Inverse Function Theorem, 93-4involute, 35-6, 285isometric, 101, 235, 296- deformation, 105-6, 216, 296isometry, 101, 111, 121, 179, 190, 229,

238-40, 297, 321-2isoperimet ric inequality, 51, 288jacobian matrix, 93, 100, 126, 160Jordan Curve Theorem, 48,252Lagrange's Method of Undeterm ined

Multipliers, 78, 292latitude, 61level curve , 2, 16, 19, 96level surface, 71limacon, 20, 51, 57, 289line of curvature, 140-1, 150, 301, 307,

309local maximum, 277-8, 280, 292, 327-8local min imum, 277-8, 280, 292, 327-8logarithmic spiral, 7, 8, 14, 34-5, 285longitude, 61loxodrome, 83lune , 118mean curvature, 147, 150, 161-2, 201,

307- constant, 203, 246Mercator's projection, 83, 111, 238

36, 175, 298156133, 137-8, 140, 162,

Index

meridian, 81, 129, 140, 182, 189, 292Meusnier's Theorem, 131Möbius band, 76-7,239-40,321Möbius transformation, 190monkey saddle, 143multiplicity, 269, 274-5, 277, 326non-euclidean geometry, 154normal- to a curve, 17- to a surface, 75normal curvature, 127, 130, 137-8, 140,

299-300normal section, 128, 173open ball, 60open disc, 60, 289open interval, 2, 60open set, 59osculating circle, 35, 285parabolic cylinder, 87parabolic point, 143, 164parallel, 81, 129, 140, 182, 189, 252,

292parallel curve, 35, 285parallel surface, 161-4parameter curve, 107, 161, 175parametrisation- of a curve, 2- of a surface, 60parametrised curve , 2, 16, 19, 21period, 47, 247planar point, 143, 164Plateau's problem, 201principal curvature, 132-3, 137-8, 140,

142-3, 148, 150, 162, 164, 244-5, 304,306

principal normal,principal patch,principal vector,

300,306profile curve, 81pseudosphere, 153-5, 185-6, 190, 235-6,

257, 304-6, 310-1quadric, 84-7, 89, 293quadric cone, 86, 295regular, 11, 15, 21regular point, 11, 16reparametrisation, 10, 69, 100, 126- unit-speed, 12, 13, 172reparametrisation map, 10, 69right conoid, 83rigid motion, 30, 42, 100, 150, 207,

242, 285-6Rodrigues' formula, 140

331

ruling, 80, 293, 296saddle point, 277-8, 280, 327-8Scherk's surface, 164, 214, 216, 317second fundamental form, 125-6, 242signed unit normal, 28, 51simple closed curve, 47, 247-8, 288- area of, 50-1- exterior of, 48- interior of, 48, 248- length of, 49, 51- period of 47, 247- positively-oriented, 49, 248simply-connected, 221sink, 271, 273smooth- curve, 4, 74, 96- function, 73, 78- map, 69- surface, 67, 71soap films, 202-3source, 271, 273speed, 9, 172spheroid, 190, 311-2standard unit normal, 75, 139stationary point, 269, 271, 274-5stereographie projection, 110, 238, 296surface, 60- area of, 113-5- compact, 164, 207, 244, 246, 258,

260, 268, 271- conjugate, 223, 226-7, 318- Hat, 155, 164- level, 71- minimal, 202, 207-8, 211, 214-7,

219-21- of revolution, 81, 89, 125, 129, 140,

145, 149, 182-3, 189, 208, 234, 257- orientable, 75, 78, 166- parallel, 161-4- ruled, 80, 89, 111, 149-50, 156, 164,

211, 304- smooth 67, 71- translation, 216, 316surface patch, 60- allowable 68-70- regular 67-8surface variation, 201-2, 206tangent developable, 103-4, 157-8tangent plane, 75, 292tangent space, 74tangent vector, 4, 5, 51tangent vector field, 269, 274, 326

332

Theorema Egregium, 229, 238, 240,242,321

third fundamental form, 141, 150, 302torsion, 37-8, 40-1, 44, 151, 175torus, 73, 144, 150, 169, 190, 258, 274,

291, 297, 312tractrix, 153transition map, 65, 68, 72, 291triangulation, 259-60, 324-5triply orthogonal system, 90, 92-3, 141,

164,294tube, 115, 175, 297twisted cubic, 14umbilic, 133, 144-5, 155, 207, 245, 302unit-speed, 9


unit sphere, 61, 65, 67-8, 71-2, 135,152, 168, 177

upper half-plane model, 155Utilities problem, 268vertex- of a curve, 55, 289- of a curvilinear polygon, 252Viviani's curve, 6, 45vortex, 271Weierstrass's representation, 224,

226-8Weingarten matrix, ·139Wirtinger's inequality, 52, 54witch of Agnesi, 6

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