DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces Preliminary Version Fall,2008 Theodore Shifrin University of Georgia Dedicated to the memory of Shiing-Shen Chern, my adviser and friend c2008 Theodore Shifrin No portion of this work may be reproduced in any form without written permission of the author
DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces Preliminary Version Fall, 2008 Theodore Shifrin University of Georgia Dedicated to the memory of Shiing-Shen Chern, my adviser and friend c 2008 Theodore Shifrin No portion of this work may be reproduced in any form without written permission of the author
CONTENTS 1.CURVES. 1.Examples,Arclength Parametrization 1 2.Local Theory:Frenet Frame 10 3.Some Global Results 22 2.SURFACES:LOCAL THEORY 35 1.Parametrized Surfaces and the First Fundamental Form 35 2.The Gauss Map and the Second Fundamental Form 44 3.The Codazzi and Gauss Equations and the Fundamental Theorem of Surface Theory 56 4.Covariant Differentiation,Parallel Translation,and Geodesics 65 3.SURFACES:FURTHER TOPICS 76 1.Holonomy and the Gauss-Bonnet Theorem 76 2.An Introduction to Hyperbolic Geometry 3.Surface Theory with Differential Forms 98 4.Calculus of Variations and Surfaces of Constant Mean Curvature 103 Appendix. REVIEW OF LINEAR ALGEBRA AND CALCULUS 110 1.Linear Algebra Review 110 2.Calculus Review 112 3.Differential Equations 114 SOLUTIONS TO SELECTED EXERCISES 117 INDEX 120 Problems to which answers or hints are given at the back of the book are marked with amental exercis ularly important(and to which November,2008
CONTENTS 1. CURVES . . . . . . . . . . . . . . . . . . . . 1 1. Examples, Arclength Parametrization 1 2. Local Theory: Frenet Frame 10 3. Some Global Results 22 2. SURFACES: LOCAL THEORY . . . . . . . . . . . . 35 1. Parametrized Surfaces and the First Fundamental Form 35 2. The Gauss Map and the Second Fundamental Form 44 3. The Codazzi and Gauss Equations and the Fundamental Theorem of Surface Theory 56 4. Covariant Differentiation, Parallel Translation, and Geodesics 65 3. SURFACES: FURTHER TOPICS . . . . . . . . . . . 76 1. Holonomy and the Gauss-Bonnet Theorem 76 2. An Introduction to Hyperbolic Geometry 88 3. Surface Theory with Differential Forms 98 4. Calculus of Variations and Surfaces of Constant Mean Curvature 103 Appendix. REVIEW OF LINEAR ALGEBRA AND CALCULUS . . . 110 1. Linear Algebra Review 110 2. Calculus Review 112 3. Differential Equations 114 SOLUTIONS TO SELECTED EXERCISES . . . . . . . 117 INDEX . . . . . . . . . . . . . . . . . . . 120 Problems to which answers or hints are given at the back of the book are marked with an asterisk (*). Fundamental exercises that are particularly important (and to which reference is made later) are marked with a sharp (] ). November, 2008
CHAPTER 1 Curves 1.Examples,Arclength Parametrization We say a vector function f(a.b)R3is (=0.1.2.)if f and its first k derivatives, ),are all continuous.We say fis smooth if fis for every positive integerk.Aparametrized curve is a e3 (or smooth)map a:IR3 for some interval I=(a.b)or [a,b]in R(possibly infinite).We say c is regular if'≠0 for all1∈L. We can imagine a particle moving along the path with its position at time given by().As we learned in vector calculus, se h is the velocity of the particle at timet.The velocity vector(t)is tangent to the curve at(and its length. (l,is the speed of the particle. Example 1.We begin with some standard examples. (a)Familiar from linear algebra and vector calculus is a parametrized line:Given points P and in all paths from P to the"straight line path"gives the shortest.This is typical of problems we shall consider in the future. (b)Essentially by the very definition of the trigonometric functions cosand sin,we obtain a very natura parametrization of a circle of radius a,as pictured in Figure 1.1(a): a)=a(cost,sin）=(a cost,a sint),0≤t≤2 r cos t a sin os t.b sin 1) (a) 6 FIGURE 1.1
CHAPTER 1 Curves 1. Examples, Arclength Parametrization We say a vector function fW.a; b/ ! R 3 is C k (k D 0; 1; 2; : : :) if f and its first k derivatives, f 0 , f 00 , . . . , f .k/, are all continuous. We say f is smooth if f is C k for every positive integer k. A parametrized curve is a C 3 (or smooth) map ˛W I ! R 3 for some interval I D .a; b/ or Œa; b in R (possibly infinite). We say ˛ is regular if ˛ 0 .t / ¤ 0 for all t 2 I . We can imagine a particle moving along the path ˛, with its position at time t given by ˛.t /. As we learned in vector calculus, ˛ 0 .t / D d˛ dt D lim h!0 ˛.t C h/ ˛.t / h is the velocity of the particle at time t. The velocity vector ˛ 0 .t / is tangent to the curve at ˛.t / and its length, k˛ 0 .t /k, is the speed of the particle. Example 1. We begin with some standard examples. (a) Familiar from linear algebra and vector calculus is a parametrized line: Given points P and Q in R 3 , we let v D ! PQ D Q P and set ˛.t / D P C tv, t 2 R. Note that ˛.0/ D P, ˛.1/ D Q, and for 0 t 1, ˛.t / is on the line segment PQ. We ask the reader to check in Exercise 8 that of all paths from P to Q, the “straight line path” ˛ gives the shortest. This is typical of problems we shall consider in the future. (b) Essentially by the very definition of the trigonometric functions cos and sin, we obtain a very natural parametrization of a circle of radius a, as pictured in Figure 1.1(a): ˛.t / D a cost;sin t D a cost; a sin t ; 0 t 2: (a cost, a sin t) (a cost, b sin t) t a a b (a) (b) FIGURE 1.1 1
CHAPTER 1.CURVES (c)Now,if a,b>0and we apply the linear map T:R2R2,T(x.y)=(ax.by). we see that the unit circlex2+y2=1maps to the ellipsex2/a2+y2/b2=1.Since T(cost.sint)= (a cost.bsin),the latter gives a natural parametrization of the ellipse,as shown in Figure 1.1(b). (d)Consider the two cubic curves inR2 illustrated in Figure 1.2.On the left is the cuspidal cubi (a) b FIGURE 1.2 y2=x3,and on the right is the nodal cubicy2=x+x2.These can be parametrized,respectively by the functions a)=(2,3)and a0=2-1,t2-1) (In the latter case,as the figure suggests,we see that the line y=tx intersects the curve when x)2=x2x+10,s0x=0orx=2-1)
2 CHAPTER 1. CURVES (c) Now, if a; b > 0 and we apply the linear map T W R 2 ! R 2 ; T .x; y/ D .ax; by/; we see that the unit circle x 2Cy 2 D 1 maps to the ellipse x 2=a2Cy 2=b2 D 1. Since T .cost;sin t / D .a cost; b sin t /, the latter gives a natural parametrization of the ellipse, as shown in Figure 1.1(b). (d) Consider the two cubic curves in R 2 illustrated in Figure 1.2. On the left is the cuspidal cubic y=tx y 2=x 3 y 2=x 3+x 2 (a) (b) FIGURE 1.2 y 2 D x 3 , and on the right is the nodal cubic y 2 D x 3Cx 2 . These can be parametrized, respectively, by the functions ˛.t / D .t2 ; t3 / and ˛.t / D .t2 1; t .t2 1//: (In the latter case, as the figure suggests, we see that the line y D tx intersects the curve when .tx/2 D x 2 .x C 1/, so x D 0 or x D t 2 1.) z=x 3 y=x 2 z 2=y 3 FIGURE 1.3
$1.EXAMPLES,ARCLENGTH PARAMETRIZATION (e)Now consider the fwisted cubic in R3,illustrated in Figure 1.3,given by a0=,2,.teR Its projections in the xyxzand yz-coordinate planes are.respectively,y=x2z=x3,and 22=y3 (the cuspidal cubic). (f)Our next (circle).Consider the illustration in Figure 1.4.Assuming the wheel rolls without slipping.the a FIGURE 1.4 distance it travels along the ground is equal to the length of the circular are subtended by the angle through which it has tured.That is,if the radius of the cirle isand it has turned through angl t,then the point of contact with the x-axis,.is at units to the right.The vector from the origin to FIGURE 1.5 the point P can be expressed as the sum of the three vectors C,and CP(see Figure 1.5): 03=00+Q元+c =(at.0)+(0.a)+(-asint.-acost) and hence the function a(t)=(at -a sint,a-acost)=a(t-sint.I-cost).tER gives a parametrization of the cycloid. (g)A(circular)helix is the screw-like path of a bug as it walks uphill on a right circular cylinder at a constant slope or pitch.If the cylinder has radius a and the slope is b/a,we can imagine drawing a line of that slope on a piece of paper 2a units long,and then rolling the paper up into a cylinder. The line gives one revolution of the helix,as we can see in Figure 1.6.If we take the axis of the cylinder to be vertical,the projection of the helix in the horizontal plane is a circle of radius a,and so we obtain the parametrization a(t)=(a cost,a sint,br)
÷1. EXAMPLES, ARCLENGTH PARAMETRIZATION 3 (e) Now consider the twisted cubic in R 3 , illustrated in Figure 1.3, given by ˛.t / D .t; t2 ; t3 /; t 2 R: Its projections in the xy-, xz-, and yz-coordinate planes are, respectively, y D x 2 , z D x 3 , and z 2 D y 3 (the cuspidal cubic). (f) Our next example is a classic called the cycloid: It is the trajectory of a dot on a rolling wheel (circle). Consider the illustration in Figure 1.4. Assuming the wheel rolls without slipping, the t O P a FIGURE 1.4 distance it travels along the ground is equal to the length of the circular arc subtended by the angle through which it has turned. That is, if the radius of the circle is a and it has turned through angle t, then the point of contact with the x-axis, Q, is at units to the right. The vector from the origin to t a cost a sin t a P C O P Q C FIGURE 1.5 the point P can be expressed as the sum of the three vectors ! OQ, ! QC, and ! CP (see Figure 1.5): ! OP D ! OQ C ! QC C ! CP D .at; 0/ C .0; a/ C .a sin t; a cost /; and hence the function ˛.t / D .at a sin t; a a cost / D a.t sin t; 1 cost /; t 2 R gives a parametrization of the cycloid. (g) A (circular) helix is the screw-like path of a bug as it walks uphill on a right circular cylinder at a constant slope or pitch. If the cylinder has radius a and the slope is b=a, we can imagine drawing a line of that slope on a piece of paper 2a units long, and then rolling the paper up into a cylinder. The line gives one revolution of the helix, as we can see in Figure 1.6. If we take the axis of the cylinder to be vertical, the projection of the helix in the horizontal plane is a circle of radius a, and so we obtain the parametrization ˛.t / D .a cost; a sin t; bt /
CHAPTER 1.CURVES 2π1 FIGURE1.6 Brief review.Just as the circle=1 is parametrized by (cos.sin),the portion of the hyperbolax2-y2=1 lying to the right of the y-axis,as shown in Figure 1.7,is parametrized by (cosht,sinht),where cosht=e +e- and 2 sinht =ef -e- 2 sinht By analogy with circular trigonometry,we set tanht= (cosh t.sinh) FIGURE1.7 formulas are easy to check cosh2 t-sinh2 t=1. tanh2 t +scch2t=I sinh'(r)=cosht cosh'(t)=sinht tanh'(t)=sech2t. sech'(t)=-tanht secht
4 CHAPTER 1. CURVES FIGURE 1.6 Brief review of hyperbolic trigonometric functions. Just as the circle x 2Cy 2 D 1 is parametrized by .cos ;sin /, the portion of the hyperbola x 2 y 2 D 1 lying to the right of the y-axis, as shown in Figure 1.7, is parametrized by .cosh t;sinh t /, where cosh t D e t C e t 2 and sinh t D e t e t 2 : By analogy with circular trigonometry, we set tanh t D sinh t cosh t and secht D 1 cosh t . The following (cosh t, sinh t) FIGURE 1.7 formulas are easy to check: cosh2 t sinh2 t D 1; tanh2 t C sech2 t D 1 sinh0 .t / D cosh t; cosh0 .t / D sinh t; tanh0 .t / D sech2 t; sech0 .t / D tanh t secht:
$1.EXAMPLES,ARCLENGTH PARAMETRIZATION 5 (h)When a uniform and flexible chain hangs from two its length.The shape it takes is called a catenary.As we ask the reader to check in Exercise 9. the catenary is the graph of f(x)=C cosh(x/C).for any constant C>0.This curve will appear FIGURE 1.8 numerous times in this course. Example 2.One of the more interesting curves that arises"in nature"is the traditional story is this:A dog is at the end of a l-unit leash and buries a bone at (0.1)as his owner begins to walk down the x-axis,starting at the origin.The dog tries to get back to the bone,so he always pulls the leash taut as he is dragged along the tractrix by his owner.His pulling the leash taut means that the leash will be tangent to the curve.When the master is at (1.).let the dog's position be (x(r).()).and let the leash 4(0.1) (x,y) FIgUre 1.9 make angle 0(t)with the positive x-axis.Then we havex(r)=t +cos(t),y(t)=sin(t),so m0=贵-0=品 cos8()8') Therefore,0'(r)=sin().Separating variables and integrating.we have fd0/sin=fdr,and so t =-In(csc 6 cot 0)+c for some constant c.Since =/2 when t =0,we see that c =0.Now, 2c0s2(8/2) since cse +cotcicot(2),we can rewrite this asIn tan(). sin Thus,we can parametrize the tractrix by x(0)=(cos6+In tan(0/2),sin8),π/2≤0<元
÷1. EXAMPLES, ARCLENGTH PARAMETRIZATION 5 (h) When a uniform and flexible chain hangs from two pegs, its weight is uniformly distributed along its length. The shape it takes is called a catenary. 1 As we ask the reader to check in Exercise 9, the catenary is the graph of f .x/ D C cosh.x=C /, for any constant C > 0. This curve will appear FIGURE 1.8 numerous times in this course. O Example 2. One of the more interesting curves that arises “in nature” is the tractrix. 2 The traditional story is this: A dog is at the end of a 1-unit leash and buries a bone at .0; 1/ as his owner begins to walk down the x-axis, starting at the origin. The dog tries to get back to the bone, so he always pulls the leash taut as he is dragged along the tractrix by his owner. His pulling the leash taut means that the leash will be tangent to the curve. When the master is at .t; 0/, let the dog’s position be .x.t /; y.t //, and let the leash FIGURE 1.9 make angle .t / with the positive x-axis. Then we have x.t / D t C cos .t /, y.t / D sin .t /, so tan .t / D dy dx D y 0 .t / x 0 .t / D cos .t /0 .t / 1 sin .t /0 .t /: Therefore, 0 .t / D sin .t /. Separating variables and integrating, we have R d= sin D R dt, and so t D ln.csc C cot / C c for some constant c. Since D =2 when t D 0, we see that c D 0. Now, since csc Ccot D 1 C cos sin D 2 cos2 .=2/ 2 sin.=2/ cos.=2/ D cot.=2/, we can rewrite this as t D ln tan.=2/. Thus, we can parametrize the tractrix by ˛. / D cos C ln tan.=2/;sin ; =2 < : 1From the Latin catena, chain. 2From the Latin trahere, tractus, to pull
6 CHAPTER 1.CURVES Alternatively,since tan(0/2)=e,we have 2e' 2 sin0 =2sin(@/2)cos(/2)=sech e-t-er 匹0=cos02-02片ah and so we can parametrize the tractrix instead by B()=(t-tanh.secht),t≥0.7 The fundamental concept underlying the geometry of curves is the arclength of a parametrized curve. Definition.If [a,b]R3 is a parametrized curve,then for any ab,we define its arclength from a to t to be s(t)= ()du.That is,the distance a particle travels-the arclength of its trajectory-is the integral of its speed. An alternative approach is to start with the following Definition.Let:[a.b]R3 be a(continuous)parametrized curve.Given a partition P=a=to< <.<t=b)of the interval [a.b],let (a,P)=Ia(G)-at-1). i=1 That is,is the length of the inscribed polygon with vertices at ),i=0.k,as indicated in Given this partition.of] the length of this polygonal path is (o.) FIGURE 1.10 Figure 1.10.We define the arclength of to be length()=sup(.):a partition of [a.] provided the set of polygonal lengths is bounded above. Now,using this definition,we can prove that the distance a particle travels is the integral of its speed. We will need to use the result of Exercise A.2.4
6 CHAPTER 1. CURVES Alternatively, since tan.=2/ D e t , we have sin D 2 sin.=2/ cos.=2/ D 2et 1 C e 2t D 2 e t C et D secht cos D cos2 .=2/ sin2 .=2/ D 1 e 2t 1 C e 2t D e t e t e t C et D tanh t; and so we can parametrize the tractrix instead by ˇ.t / D t tanh t;secht /; t 0: O The fundamental concept underlying the geometry of curves is the arclength of a parametrized curve. Definition. If ˛W Œa; b ! R 3 is a parametrized curve, then for any a t b, we define its arclength from a to t to be s.t / D Z t a k˛ 0 .u/kdu. That is, the distance a particle travels—the arclength of its trajectory—is the integral of its speed. An alternative approach is to start with the following Definition. Let ˛W Œa; b ! R 3 be a (continuous) parametrized curve. Given a partition P D fa D t0 < t1 < < tk D bg of the interval Œa; b, let `.˛;P/ D X k iD1 k˛.ti/ ˛.ti1/k: That is, `.˛;P/ is the length of the inscribed polygon with vertices at ˛.ti/, i D 0; : : : ; k, as indicated in a b FIGURE 1.10 Figure 1.10. We define the arclength of ˛ to be length.˛/ D supf`.˛;P/ W P a partition of Œa; bg; provided the set of polygonal lengths is bounded above. Now, using this definition, we can prove that the distance a particle travels is the integral of its speed. We will need to use the result of Exercise A.2.4
$1.EXAMPLES,ARCLENGTH PARAMETRIZATION Proposition 1.1.Leta:[a,b]R3 be a piecewise-l parametrized curve.Then length(a)=l()dt. Proof.For any partitionof [a,b],we have so length(a)≤ (ldt.The same holds on any interval. Now,for b,define s(r)to be the arclength of the curve a on the interval [a.].Then forh> we have h since s(+h)-s(r)is the arelength of the curve a on the interval [(See Exercise 8 for the first inequality.)Now Therefore,by the squeeze principle, 思0+》-0=w0m h A similar argument works for h,and we conclude thats)=().Therefore s)=la'(u)ldu,a≤tsb, and,in particular,s(b)=length(@)= 。le()ldi,as desired. We say the curve a is parametrized by arclength if (=1 for all t,so that s(t)=t-a.In this event,we usually use the parameter s and write (s). Example 3. (a)The standard parametrization of the circle of radius a is ( (a cost,a sint),t [0.2].so a'(t)=(-a sint.a cost)and llo'(t)ll a.It is easy to see from the chain rule that if we reparametrize the curve by B(s)=(a cos(s/a).a sin(s/a)),s [0.2ma], then B'(s)=(-sin(s/a).cos(s/a))and B'(s)=1 for all s.Thus,the curve B is parametrized by arclength. (b)Let a(s)=((1+s)3/2.(1-s)32.s),s e(-1.1).Then we have a'(s)=((1+)12. -l-sy/2,方》，and'sl=1 for all.s.Thus is parametried by arclength
÷1. EXAMPLES, ARCLENGTH PARAMETRIZATION 7 Proposition 1.1. Let ˛W Œa; b ! R 3 be a piecewise-C 1 parametrized curve. Then length.˛/ D Z b a k˛ 0 .t /kdt: Proof. For any partition P of Œa; b, we have `.˛;P/ D X k iD1 k˛.ti/ ˛.ti1/k D X k iD1 Z ti ti1 ˛ 0 .t /dt X k iD1 Z ti ti1 k˛ 0 .t /kdt D Z b a k˛ 0 .t /kdt; so length.˛/ Z b a k˛ 0 .t /kdt. The same holds on any interval. Now, for a t b, define s.t / to be the arclength of the curve ˛ on the interval Œa; t. Then for h > 0 we have k˛.t C h/ ˛.t /k h s.t C h/ s.t / h 1 h Z tCh t k˛ 0 .u/kdu; since s.t C h/ s.t / is the arclength of the curve ˛ on the interval Œt; t C h. (See Exercise 8 for the first inequality.) Now lim h!0C k˛.t C h/ ˛.t /k h D k˛ 0 .t /k D lim h!0C 1 h Z tCh t k˛ 0 .u/kdu: Therefore, by the squeeze principle, lim h!0C s.t C h/ s.t / h D k˛ 0 .t /k: A similar argument works for h < 0, and we conclude that s 0 .t / D k˛ 0 .t /k. Therefore, s.t / D Z t a k˛ 0 .u/kdu; a t b; and, in particular, s.b/ D length.˛/ D Z b a k˛ 0 .t /kdt, as desired. We say the curve ˛ is parametrized by arclength if k˛ 0 .t /k D 1 for all t, so that s.t / D t a. In this event, we usually use the parameter s and write ˛.s/. Example 3. (a) The standard parametrization of the circle of radius a is ˛.t / D .a cost; a sin t /, t 2 Œ0; 2, so ˛ 0 .t / D .a sin t; a cost / and k˛ 0 .t /k D a. It is easy to see from the chain rule that if we reparametrize the curve by ˇ.s/ D .a cos.s=a/; a sin.s=a//, s 2 Œ0; 2a, then ˇ 0 .s/ D . sin.s=a/; cos.s=a// and kˇ 0 .s/k D 1 for all s. Thus, the curve ˇ is parametrized by arclength. (b) Let ˛.s/ D 1 3 .1 C s/3=2 ; 1 3 .1 s/3=2 ; p 1 2 s , s 2 .1; 1/. Then we have ˛ 0 .s/ D 1 2 .1 C s/1=2 ; 1 2 .1 s/1=2 ; p 1 2 , and k˛ 0 .s/k D 1 for all s. Thus, ˛ is parametrized by arclength. O
CHAPTER 1.CURVES An important observation from a theoretical standpoint is that any regular parametrized curve can be reparametrized by arclength.For if is regular,the arclength functions()= l(u)ldu is an increas- ing function (since s'()=(0for all )and therefore has an inverse function=(s).Then we can consider the parametrization B(s)=a(t(s)). Note that the chain rule tells us that B'(s)=a(t(s))t'(s)=a'(t(s))/s'(t(5))=a'(t(s))/lla'(t(s))Il is everywhere a unit vector,in other words,B moves with speed 1. EXERCISES 1.1 *1.Parametrize the unit circle(less the point (-1.0))by the lengtht indicated in Figure 1.11. (-1,0の FIGURE 1.11 Consider the helix()=(acost.asint.br).Calculate).and reparametrize by ar- clength. 3.Let0）=(3os1+方sin.方eos.方cos-方sn).Calculateo,lal. by arclength *4.Parametrize the graph y=f(x),a sxsb,and show that its arclength is given by the traditional formula 5.a.Show that the arclength of the catenary)=(亿，cosh)for0≤t≤is sinh b. b.Reparametrize the catenary by arclength.(Hint:Find the inverse of sinh by using the quadratic formula.) 6.Consider the curve()=(e.e-)Calculate a(),(),and reparametrize by arclength, starting at =0
8 CHAPTER 1. CURVES An important observation from a theoretical standpoint is that any regular parametrized curve can be reparametrized by arclength. For if ˛ is regular, the arclength function s.t / D Z t a k˛ 0 .u/kdu is an increasing function (since s 0 .t / D k˛ 0 .t /k > 0 for all t), and therefore has an inverse function t D t .s/. Then we can consider the parametrization ˇ.s/ D ˛.t .s//: Note that the chain rule tells us that ˇ 0 .s/ D ˛ 0 .t .s//t0 .s/ D ˛ 0 .t .s//=s0 .t .s// D ˛ 0 .t .s//=k˛ 0 .t .s//k is everywhere a unit vector; in other words, ˇ moves with speed 1. EXERCISES 1.1 *1. Parametrize the unit circle (less the point .1; 0/) by the length t indicated in Figure 1.11. t (−1,0) (x,y) FIGURE 1.11 ]2. Consider the helix ˛.t / D .a cost; a sin t; bt /. Calculate ˛ 0 .t /, k˛ 0 .t /k, and reparametrize ˛ by arclength. 3. Let ˛.t / D p 1 3 costCp 1 2 sin t; p 1 3 cost; p 1 3 costp 1 2 sin t . Calculate ˛ 0 .t /, k˛ 0 .t /k, and reparametrize ˛ by arclength. *4. Parametrize the graph y D f .x/, a x b, and show that its arclength is given by the traditional formula length D Z b a q 1 C f 0 .x/2 dx: 5. a. Show that the arclength of the catenary ˛.t / D .t; cosh t / for 0 t b is sinh b. b. Reparametrize the catenary by arclength. (Hint: Find the inverse of sinh by using the quadratic formula.) *6. Consider the curve ˛.t / D .et ; et ; p 2t /. Calculate ˛ 0 .t /, k˛ 0 .t /k, and reparametrize ˛ by arclength, starting at t D 0